= Solution
Write $u_i=u^{s_i}(p_i)$ and $v_i=v^{s_i}(p_i)$. With $S_{fi}=\delta_{fi}+i(2\pi)^4\delta^{(4)}(p_1+p_2-p_3-p_4)\mathcal M_{fi}$, choose
$$
\boxed{\mathcal M_{p_1,p_2,p_3,p_4}^{s_1,s_2,s_3,s_4}=g^2(T-A),\qquad
T=(\bar u_3u_1)(\bar v_2v_4),\quad A=(\bar v_2u_1)(\bar u_3v_4).}
$$
The relative minus sign is fixed by the <canonical anticommutation relations>, not by the <Mandelstam variables>. It follows by <crossing symmetry> from the identical-<fermion> <scattering amplitude> in <2/b/i>, or directly by the two inequivalent external contractions of $g^2(\bar\psi\psi)^2/2$. For an explicit sign check, fix incoming state order $b_1^\dagger d_2^\dagger|0\rangle$ and outgoing order $b_3^\dagger d_4^\dagger|0\rangle$. The direct contraction contains $(b_3^\dagger b_1)(d_2d_4^\dagger)$, whose coefficient in the ordered product $b_3^\dagger d_4^\dagger d_2b_1$ is $-1$; the annihilation contraction $(b_3^\dagger d_4^\dagger)(d_2b_1)$ has coefficient $+1$. The common sign can change with the external-state convention, while their relative minus sign cannot.
Let $r=m^2$. The external particles are <on shell>, so $s+t+u=4r$ and
$$
p_1\cdot p_2=p_3\cdot p_4=\frac s2-r,\qquad
p_1\cdot p_3=p_2\cdot p_4=r-\frac t2,\qquad
p_1\cdot p_4=p_2\cdot p_3=r-\frac u2.
$$
Use the supplied <spin sums> and <gamma matrix trace identities>. The two diagonal terms in the initial <spin average> are
$$
\frac14\sum|T|^2
=\frac14\operatorname{Tr}[(\not p_3+m)(\not p_1+m)]
\operatorname{Tr}[(\not p_2-m)(\not p_4-m)]
=(t-4r)^2,
$$
$$
\frac14\sum|A|^2
=\frac14\operatorname{Tr}[(\not p_2-m)(\not p_1+m)]
\operatorname{Tr}[(\not p_3+m)(\not p_4-m)]
=(s-4r)^2.
$$
For the interference term, the <spin sum> is a single <Dirac trace>:
$$
R:=\sum TA^*
=\operatorname{Tr}[(\not p_3+m)(\not p_1+m)(\not p_2-m)(\not p_4-m)].
$$
Terms with an odd number of <Dirac gamma matrices> vanish. Expanding the remaining terms gives
$$
\begin{aligned}
\frac R4={}&(p_3\cdot p_1)(p_2\cdot p_4)
-(p_3\cdot p_2)(p_1\cdot p_4)
+(p_3\cdot p_4)(p_1\cdot p_2)\\
&+r(p_3\cdot p_1-p_3\cdot p_2-p_3\cdot p_4-p_1\cdot p_2-p_1\cdot p_4+p_2\cdot p_4)+r^2.
\end{aligned}
$$
Substituting the <Mandelstam variables> reduces this to
$$
R=s^2+t^2-u^2+16ru-16r^2
=(s-4r)^2+(t-4r)^2-(u-4r)^2.
$$
It is real. The initial <spin average> therefore yields
$$
\overline{|\mathcal M|^2}=g^4\left[(t-4r)^2+(s-4r)^2-\frac R2\right],
$$
$$
\boxed{\overline{|\mathcal M|^2}=\frac{g^4}{2}\left[(s-4m^2)^2+(t-4m^2)^2+(u-4m^2)^2\right].}
$$
\b[The displayed answer in the source is false.] At threshold, $p_i=(m,\mathbf0)$, hence $s=4m^2$ and $t=u=0$. Then $\bar vu=\bar uv=0$, while the <spin sums> give $\overline{|\mathcal M|^2}=16g^4m^4$. The printed expression instead gives $20g^4m^4$. This is also a counterexample in the continuous threshold limit from physical nonzero incoming <momenta>. At $m=0$ there is a further discrepancy: the correct result is $g^4(s^2+t^2+u^2)/2$, rather than the printed $g^4(3s^2+3t^2-u^2)/2$. Changing a common phase of the <scattering amplitude> cannot change either comparison.
The requested $s\leftrightarrow t$ symmetry nevertheless holds for the correct answer. <Crossing symmetry> exchanges the two ways of pairing the four external <fermion> lines; the momentum-independent <auxiliary-field propagator> supplies no channel-dependent denominator. The relative <fermionic sign> can reverse the crossed <scattering amplitude>, but disappears in its squared modulus. With $u=4m^2-s-t$, the boxed expression is manifestly symmetric under $s\leftrightarrow t$. This is an analytic crossing relation, not a statement that positive physical $s$ and negative physical $t$ have interchangeable scattering kinematics.
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