= Solution
In the <laboratory frame>, the initial target is at rest and the outgoing directions are directly related to detector positions. The <centre-of-momentum frame> has zero total <momentum>, so the two outgoing <momenta> are equal and opposite; it cleanly separates bulk motion from relative <kinetic energy>. Transforming back is needed to predict laboratory angles and <speeds>. Neither <reference frame> removes the need for an appropriate collision law.
Take the incident direction as $\widehat{\mathbf x}$ and put $S=M_1+M_2$. <Momentum conservation> fixes the <centre-of-momentum frame> <velocity>, and a <Galilean transformation> gives the initial <velocities> in that frame:
$$
\boxed{\mathbf V=\frac{M_1u_1}{S}\widehat{\mathbf x},\qquad
\mathbf w_1=\frac{M_2u_1}{S}\widehat{\mathbf x},\qquad
\mathbf w_2=-\frac{M_1u_1}{S}\widehat{\mathbf x}.}
$$
After the <collision>, $M_1\mathbf w'_1+M_2\mathbf w'_2=0$. Hence both particles have the same final-to-initial <speed> ratio,
$$
q=\frac{|\mathbf w'_1|}{|\mathbf w_1|}
=\frac{|\mathbf w'_2|}{|\mathbf w_2|}
=\frac{|\mathbf v'_1-\mathbf v'_2|}{u_1}.
$$
The relative <kinetic energy> is multiplied by $q^2$. This common ratio exists independently of how the outgoing direction is selected.
\b[The meaning of restitution matters here.] The usual <coefficient of restitution> $e_n$ is a ratio of normal separation and approach <speeds>, not total <speeds>. In a smooth <oblique collision>, the tangential relative <velocity> is unchanged. If the incoming relative <velocity> makes angle $\chi$ to the contact normal, then
$$
\boxed{q^2=e_n^2\cos^2\chi+\sin^2\chi.}
$$
For example, $e_n=1/2$ and $\chi=\pi/4$ give $q=\sqrt{5/8}\ne1/2$. Therefore the requested identification is not generally true with the standard normal definition. The printed equation is valid if its $e$ denotes the <relative speed retention in an oblique collision>, namely $e=q$, or if an additional collision model supplies that equality. The elastic cases have $q=e_n=1$ and are unaffected by this qualification.
To derive the scattering formula under that convention, write $\mathbf v'_1=\mathbf V+\mathbf w'_1$. Its components give the <laboratory scattering circle>
$$
(v_1\cos\theta_1-V)^2+(v_1\sin\theta_1)^2
=q^2\left(\frac{M_2u_1}{S}\right)^2.
$$
Solving the resulting quadratic for the nonnegative <speed> yields
$$
\boxed{\frac{v_1}{u_1}=\frac{M_1}{S}\left[\cos\theta_1\pm
\sqrt{\cos^2\theta_1-1+q^2\left(\frac{M_2}{M_1}\right)^2}\right].}
$$
With $q=e$, this is the requested formula. A real square root and $v_1\ge0$ are both necessary: a negative algebraic root is not a second physical <speed>, and $v_1=0$ does not define a <scattering angle>. The subsequent limits are taken using this <laboratory scattering circle>.
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