= Solution
Assume unpolarized beams, $s\gg m_\mu^2$, a single exchanged <photon>, full azimuthal coverage and no radiative corrections. Take $c_*=\cos\theta^*$ between incoming $e^-$ and outgoing $\mu^-$. The four nonzero <helicity amplitudes> have magnitudes $e^2(1+c_*)$ for equal incoming/outgoing negative-particle <helicities>, and $e^2(1-c_*)$ for opposite ones. Averaging four initial spin states and summing the final ones gives
$$
\overline{|\mathcal M|^2}=\frac{e^4}{4}\{2(1+c_*)^2+2(1-c_*)^2\}=e^4(1+c_*^2).
$$
The massless two-body <phase space> gives
$$
\frac{d\sigma}{d\Omega_*}=\frac{\alpha^2}{4s}(1+c_*^2),\qquad
\frac{d\sigma}{dc_*}=\frac{\pi\alpha^2}{2s}(1+c_*^2),\qquad
\sigma=\frac{4\pi\alpha^2}{3s}.
$$
The even distribution reflects the equal left/right <photon> couplings and is symmetric under interchange of forward and backward directions.
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