= Solution
Write $\Gamma=(1-v^2/c^2)^{-1/2}$ for the <Lorentz factor> of the <inertial frame> $S'$, distinguishing it from the particles' changing <Lorentz factors>. Apply the <Lorentz transformation>
$$
x'=\Gamma(x-vt),\qquad t'=\Gamma(t-vx/c^2)
$$
to each <world line> in <1/a>. Both particles initially have <velocity> $-v$ in $S'$. The switching events have coordinates
$$
(t'_e,x'_e)=(0,0),\qquad
(t'_p,x'_p)=(-\Gamma vd/c^2,\Gamma d).
$$
Thus the <Positron> starts accelerating earlier in $S'$. Its <velocity> first becomes less negative, vanishes when its laboratory <velocity> reaches $v$, and then becomes positive. The <electron> continues at $-v$ until $t'=0$, then accelerates toward $-c$. The <relativistic velocity-addition formula> makes this precise:
$$
u'_e=\frac{u_e-v}{1-vu_e/c^2},\qquad
u'_p=\frac{u_p-v}{1-vu_p/c^2}.
$$
The <electric field> parallel to the <Lorentz boost> has unchanged magnitude, but its switching surface is $t'+vx'/c^2=0$, rather than $t'=0$ everywhere.
In the laboratory, the total <three-momentum> is zero for all $t$: the two <Lorentz forces> cancel at equal laboratory times. In $S'$, each active force is still $qE$, since this is one-dimensional motion with a parallel <electric field>. Define $t'_p=-\Gamma vd/c^2$. For $v>0$ the total <three-momentum>, evaluated at \b[equal $t'$], is
$$
\boxed{P'_x(t')=
\begin{cases}
-2\Gamma mv,&t'<t'_p,\\
-2\Gamma mv+F(t'-t'_p),&t'_p\leq t'<0,\\
-2\Gamma mv+\Gamma Fvd/c^2,&t'\geq0.
\end{cases}}
$$
This follows by integrating $dp'_p/dt'=F$ from its switching event and $dp'_e/dt'=-F$ from its later switching event. During the interval between them only the <Positron> feels the prescribed <electric field>, so <momentum conservation> for the particle pair fails. Once both forces act, the total <three-momentum> is constant again.
The <equal-time momentum of an externally driven particle pair> illustrates a distinction. The sum of particle <four-momenta> \b[at a fixed pair of events] is always a <four-vector>. However, a fresh sum at equal time in each <inertial frame> generally uses different events. Here transforming an equal-$t$ laboratory sum would give $-2\Gamma v\sqrt{m^2c^4+F^2c^2t^2}/c^2$, which varies with $t$; this differs from the equal-$t'$ answer above. For example, with $c=m=F=d=1$ and $v=0.6$, the late-time equal-$t'$ sum is $-0.75$, whereas transforming the equal-$t=0$ sum gives $-1.5$. This is a concrete failure of treating the pair's equal-time sums as components of a single frame-independent <four-vector>.
A closed system's total <four-momentum>, including the <electromagnetic field> and the apparatus supplying it, is conserved and independent of the choice of spacelike integration surface under the usual isolated-system boundary conditions. The particles alone exchange <energy> and <momentum> with an external system, so neither that independence nor universal particle-only <momentum conservation> is guaranteed.
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