= Solution
The printed hint with unchanged $\sigma$ is not a valid <change of variables>: at fixed $\sigma$, the outgoing velocities forget the direction of $v-v_*$. A correct proof uses the <angular exchange for elastic collisions>.
Set $C=(v+v_*)/2$ and $v-v_*=r\omega$, where $r>0$ and $\omega\in\mathbb S^2$. Then $dv\,dv_*=dC\,r^2dr\,d\omega$, while $v'=C+r\sigma/2$ and $v_*'=C-r\sigma/2$. The gain <integral> becomes
$$
\int f(C+r\sigma/2)f(C-r\sigma/2)e^{-i(C+r\omega/2)\cdot\xi}
\,dC\,r^2dr\,d\omega\,d\sigma.
$$
Exchange the two independently integrated angular variables $\omega$ and $\sigma$. The measure is unchanged, and reverting to $v=C+r\omega/2$, $v_*=C-r\omega/2$ gives
$$
\boxed{\int f(v')f(v_*')e^{-iv\cdot\xi}\,dv\,dv_*\,d\sigma
=\int f(v)f(v_*)e^{-i(v+v_*)\cdot\xi/2}e^{-i|v-v_*|\sigma\cdot\xi/2}\,dv\,dv_*\,d\sigma}.
$$
The measure-zero set $r=0$ causes no difficulty. This proves the requested identity without invoking the false fixed-$\sigma$ Jacobian assertion.
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