You have two incoming fast momenta (\(P_{1}\) and \(P_{2}\)) and two outgoing fast momenta (\(P_{3}\) and \(P_{4}\)). This problem does not have a non-relativistic limit. I will assume elasticity:
$$ \Vert P_{1} \Vert^{2} = \Vert P_{3} \Vert^{2}, \qquad \Vert P_{2} \Vert^{2} = \Vert P_{4} \Vert^{2}. $$
Energy-Momentum Vectors
The four external fast momenta can be written as
$$ P_{1} = \begin{pmatrix} w_{1} \sinh{(\rho_{1})} & w_{1} \cosh{(\rho_{1})} \end{pmatrix}; $$
$$ P_{2} = \begin{pmatrix} w_{2} \sinh{(\rho_{2})} & w_{2} \cosh{(\rho_{2})} \end{pmatrix}; $$
$$ P_{3} = \begin{pmatrix} w_{1} \sinh{(\rho_{3})} & w_{1} \cosh{(\rho_{3})} \end{pmatrix}; $$
$$ P_{4} = \begin{pmatrix} w_{2} \sinh{(\rho_{4})} & w_{2} \cosh{(\rho_{4})} \end{pmatrix}. $$
Conservation
The conservation of energy leads to
$$ w_{1} \sinh{(\rho_{1})} + w_{2} \sinh{(\rho_{2})} = w_{1} \sinh{(\rho_{3})} + w_{2} \sinh{(\rho_{4})}. $$
Similarly, the conservation of momentum leads to
$$ w_{1} \cosh{(\rho_{1})} + w_{2} \cosh{(\rho_{2})} = w_{1} \cosh{(\rho_{3})} + w_{2} \cosh{(\rho_{4})}. $$
These two expressions can be combined to yield two equivalent expressions:
$$ w_{1} \exp{(\rho_{1})} + w_{2} \exp{(\rho_{2})} = w_{1} \exp{(\rho_{3})} + w_{2} \exp{(\rho_{4})}; $$
$$ \frac{w_{1}}{\exp{(\rho_{1})}} + \frac{w_{2}}{\exp{(\rho_{2})}} = \frac{w_{1}}{\exp{(\rho_{3})}} + \frac{w_{2}}{\exp{(\rho_{4})}}. $$
2-Mandelstam Invariants
You can introduce three 2-Mandelstam invariants:
$$ s = -(P_{1} + P_{2})^{2} = -(P_{3} + P_{4})^{2}; $$
$$ t = -(P_{1} - P_{3})^{2} = -(P_{2} - P_{4})^{2}; $$
$$ u = -(P_{1} - P_{4})^{2} = -(P_{2} - P_{3})^{2}. $$
Due to the conservation laws, you have
$$ s + t + u = -2 w_{1}^{2} - 2 w_{2}^{2}. $$
Regge-Mandelstam Invariants
You can introduce six Regge-Mandelstam invariants:
$$ r_{12} \equiv \frac{P_{1} \cdot P_{2}}{w_{1} w_{2}} = -\frac{s + w_{1}^{2} + w_{2}^{2}}{2 w_{1} w_{2}} = \cosh{(\rho_{1} - \rho_{2})}; $$
$$ r_{34} \equiv \frac{P_{3} \cdot P_{4}}{w_{1} w_{2}} = -\frac{s + w_{1}^{2} + w_{2}^{2}}{2 w_{1} w_{2}} = \cosh{(\rho_{3} - \rho_{4})}; $$
$$ r_{13} \equiv \frac{P_{1} \cdot P_{3}}{w_{1}^{2}} = \frac{t + 2w_{1}^{2}}{2 w_{1}^{2}} = \cosh{(\rho_{1} - \rho_{3})}; $$
$$ r_{24} \equiv \frac{P_{2} \cdot P_{4}}{w_{2}^{2}} = \frac{t + 2w_{2}^{2}}{2 w_{2}^{2}} = \cosh{(\rho_{2} - \rho_{4})}; $$
$$ r_{14} \equiv \frac{P_{1} \cdot P_{4}}{w_{1} w_{2}} = \frac{u + w_{1}^{2} + w_{2}^{2}}{2 w_{1} w_{2}} = \cosh{(\rho_{1} - \rho_{4})}; $$
$$ r_{23} \equiv \frac{P_{2} \cdot P_{3}}{w_{1} w_{2}} = \frac{u + w_{1}^{2} + w_{2}^{2}}{2 w_{1} w_{2}} = \cosh{(\rho_{2} - \rho_{3})}. $$
Since the hyperbolic cosine satisfies
$$ \cosh{(x)} \geq 1; $$
you have the conditions:
$$ s \leq -(w_{1} + w_{2})^{2}, \qquad t \geq 0, \qquad u \geq -(w_{1} - w_{2})^{2}. $$
These conditions define the range of values allowed for the 2-Mandelstam invariants. The first one says that \(s\) only takes negative values below the threshold value \(-(w_{1} + w_{2})^{2}\). Thus, the vector \(P_{s} = P_{1} + P_{2} = P_{3} + P_{4}\) is fast and you can write
$$ P_{s} = \begin{pmatrix} \sqrt{-s} \sinh{(\rho_{s})} & \sqrt{-s} \cosh{(\rho_{s})} \end{pmatrix}. $$
The second condition says that \(t\) only takes negative values. This means that the vector \(P_{t} = P_{1} - P_{3} = P_{4} - P_{2}\) is slow and you can write:
$$ P_{t} = \begin{pmatrix} \sqrt{t} \cosh{(\xi_{t})} & \sqrt{t} \sinh{(\xi_{t})} \end{pmatrix}. $$
Note that
$$ P_{s} \cdot P_{t} = -\sqrt{-s} \sqrt{t} \sinh{(\rho_{s} - \xi_{t})}. $$
When \(\rho_{s} = 0\) and \(\xi_{t} = 0\), you have the center-of-energy frame.
Since \(r_{12} = r_{34}\), you have
$$ \cosh{(\rho_{1} - \rho_{2})} = \cosh{(\rho_{3} - \rho_{4})}. $$
This has two solutions:
$$ \rho_{1} - \rho_{2} = \rho_{3} - \rho_{4}; $$
or
$$ \rho_{1} - \rho_{2} = \rho_{4} - \rho_{3}. $$
The first solution implies that
$$ \rho_{3} = \rho_{1} - \rho_{2} + \rho_{4}. $$
Combining this with the conservation law gives
$$ \rho_{3} = \rho_{1}, \qquad \rho_{4} = \rho_{2}. $$
This is the identity solution. The second solution is compatible with \(r_{14} = r_{23}\). Indeed, since you can write the fast rapidity in terms of the fast speed as
$$ \rho = \operatorname{acoth}{(v)}, $$
it follows that
$$ \operatorname{acoth}{(v_{1})} - \operatorname{acoth}{(v_{2})} = \operatorname{acoth}{(v_{4})} - \operatorname{acoth}{(v_{3})}. $$
Although \(r_{13} \neq r_{24}\), you have
$$ t = 2 w_{1}^{2} (1 + r_{13}) = 2 w_{2}^{2} (1 + r_{24}). $$
This means that
$$ w_{1}^{2} (1 + \cosh{(\rho_{1} - \rho_{3})}) = w_{2}^{2} (1 + \cosh{(\rho_{2} - \rho_{4})}). $$
But recall that
$$ \cosh^{2}{\left( \frac{x}{2} \right)} = \frac{\cosh{(x)} + 1}{2}. $$
Thus,
$$ w_{1}^{2} \cosh^{2}{\left( \frac{\rho_{1} - \rho_{3}}{2} \right)} = w_{2}^{2} \cosh^{2}{\left( \frac{\rho_{2} - \rho_{4}}{2} \right)}. $$
Solving for the Outgoing Data
You can first write
$$ \rho_{4} = \rho_{1} - \rho_{2} + \rho_{3}. $$
Then it follows that
$$ \exp{(\rho_{4})} = \frac{\exp{(\rho_{1})} \exp{(\rho_{3})}}{\exp{(\rho_{2})}}. $$
From the conservation law you find
$$ w_{1} \exp{(\rho_{1})} + w_{2} \exp{(\rho_{2})} = w_{1} \exp{(\rho_{3})} + w_{2} \left( \frac{\exp{(\rho_{1})} \exp{(\rho_{3})}}{\exp{(\rho_{2})}} \right). $$
This gives
$$ \exp{(\rho_{3})} = \left[\frac{ w_{1} \exp{(\rho_{1})} + w_{2} \exp{(\rho_{2})} }{w_{1} \exp{(\rho_{2})} + w_{2} \exp{(\rho_{1})} }\right] \exp{(\rho_{2})}. $$
Plugin this back into the first expression gives
$$ \exp{(\rho_{4})} = \left[\frac{ w_{1} \exp{(\rho_{1})} + w_{2} \exp{(\rho_{2})} }{w_{1} \exp{(\rho_{2})} + w_{2} \exp{(\rho_{1})} }\right] \exp{(\rho_{1})}. $$
The speeds can be introduced back via the relation
$$ v = \coth{(\rho)} \quad \Longrightarrow \quad \exp{(2 \rho)} = \frac{v + 1}{v - 1}. $$
You find
$$ \frac{\sqrt{v_{3} + 1}}{\sqrt{v_{3} - 1}} = \frac{\sqrt{v_{2} + 1}}{\sqrt{v_{2} - 1}} \frac{w_{1} \sqrt{(v_{1} + 1)(v_{2} - 1)} + w_{2} \sqrt{(v_{1} - 1)(v_{2} + 1)}}{w_{1} \sqrt{(v_{1} - 1)(v_{2} + 1)} + w_{2} \sqrt{(v_{1} + 1)(v_{2} - 1)}}; $$
$$ \frac{\sqrt{v_{4} + 1}}{\sqrt{v_{4} - 1}} = \frac{\sqrt{v_{1} + 1}}{\sqrt{v_{1} - 1}} \frac{w_{1} \sqrt{(v_{1} + 1)(v_{2} - 1)} + w_{2} \sqrt{(v_{1} - 1)(v_{2} + 1)}}{w_{1} \sqrt{(v_{1} - 1)(v_{2} + 1)} + w_{2} \sqrt{(v_{1} + 1)(v_{2} - 1)}}. $$
This leads to the identity
$$ \frac{(v_{3} + 1)(v_{4} - 1)}{(v_{3} - 1)(v_{4} + 1)} = \frac{(v_{2} + 1)(v_{1} - 1)}{(v_{2} - 1)(v_{1} + 1)}. $$
When \(w_{2} = w_{1}\), this solution reduces to
$$ \rho_{3} = \rho_{2}, \qquad \rho_{4} = \rho_{1}. $$
This is the permutation solution.
Composite Rapidities
The fast rapidity \(\rho_{s}\) associated to \(P_{s}\) can be writen as
$$ \rho_{s} = \log{\left( \frac{w_{1} \exp{(\rho_{1})} + w_{2} \exp{(\rho_{2})}}{\sqrt{-s}} \right)} = \log{\left( \frac{w_{1} \exp{(\rho_{3})} + w_{2} \exp{(\rho_{4})}}{\sqrt{-s}} \right)}. $$
Similarly, the slow rapidity \(\xi_{t}\) associated to \(P_{t}\) can be written as
$$ \xi_{t} = \log{\left( \frac{w_{1} \exp{(\rho_{1})} - w_{1} \exp{(\rho_{3})}}{\sqrt{t}} \right)} = \log{\left( \frac{w_{2} \exp{(\rho_{4})} - w_{2} \exp{(\rho_{2})}}{\sqrt{t}} \right)}. $$
Regimes
The condition on \(u\) does not allow for a specific kind of rapidity for the vector \(P_{u} = P_{1} - P_{4} = P_{3} - P_{2}\). Depending on the values of \(s\) and \(t\), you can have all three cases.
Fast \(u\)-channel
In the region
$$ -(w_{1} - w_{2})^{2} < u < 0 $$
the vector \(P_{u}\) is fast and given by
$$ P_{u} = \begin{pmatrix} \sqrt{-u} \sinh{(\rho_{u})} & \sqrt{-u} \cosh{(\rho_{u})} \end{pmatrix}. $$
Then
$$ P_{s} \cdot P_{u} = \sqrt{-s} \sqrt{-u} \cosh{(\rho_{s} - \rho_{u})}; $$
and
$$ P_{t} \cdot P_{u} = \sqrt{t} \sqrt{-u} \sinh{(\xi_{t} - \rho_{u})}. $$
The composite slow rapidity \(\rho_{u}\) is given by
$$ \rho_{u} = \log{\left( \frac{w_{1} \exp{(\rho_{1})} - w_{2} \exp{(\rho_{4})}}{\sqrt{u}} \right)} = \log{\left( \frac{w_{1} \exp{(\rho_{3})} - w_{2} \exp{(\rho_{2})}}{\sqrt{u}} \right)}. $$
Null \(u\)-channel
If \(u = 0\), then the vector \(P_{u}\) is null and can be written as
$$ P_{u} = \begin{pmatrix} k_{u} & k_{u} \end{pmatrix}. $$
Then
$$ P_{s} \cdot P_{u} = \sqrt{s} k_{u} \exp{(-\rho_{s})}; $$
and
$$ P_{t} \cdot P_{u} = \sqrt{-t} k_{u} \exp{(\rho_{t})}. $$
You also have
$$ k_{u} = w_{1} \sinh{(\rho_{1})} - w_{2} \sinh{(\rho_{4})} = w_{1} \cosh{(\rho_{1})} - w_{2} \cosh{(\rho_{4})}. $$
From here it follows that
$$ \rho_{4} = \rho_{1} - \log{\left( \frac{w_{1}}{w_{2}} \right)}. $$
Using the conservation law, you can also write
$$ k_{u} = w_{1} \sinh{(\rho_{3})} - w_{2} \sinh{(\rho_{2})} = w_{1} \cosh{(\rho_{3})} - w_{2} \cosh{(\rho_{2})}. $$
From here it follows that
$$ \rho_{3} = \rho_{2} + \log{\left( \frac{w_{1}}{w_{2}} \right)}. $$
Note that
$$ \rho_{1} + \rho_{2} = \rho_{3} + \rho_{4}. $$
Fast \(u\)-channel
If \(u < 0\), then the vector \(P_{u}\) is fast and given by
$$ P_{u} = \begin{pmatrix} \sqrt{-u} \cosh{(\rho_{u})} & \sqrt{-u} \sinh{(\rho_{u})} \end{pmatrix}. $$
Then
$$ P_{s} \cdot P_{u} = \sqrt{s} \sqrt{-u} \cosh{(\rho_{s} - \rho_{u})}; $$
and
$$ P_{t} \cdot P_{u} = \sqrt{-t} \sqrt{-u} \sinh{(\rho_{t} - \rho_{u})}. $$
The composite fast rapidity \(\rho_{u}\) can be written as
$$ \rho_{u} = \log{\left( \frac{w_{1} \exp{(\rho_{1})} - w_{2} \exp{(\rho_{4})}}{\sqrt{-u}} \right)} = \log{\left( \frac{w_{1} \exp{(\rho_{3})} - w_{2} \exp{(\rho_{2})}}{\sqrt{-u}} \right)}. $$
