= Opposite currents
Keep wire 1's <Electrons> moving at $+v$, but let wire 2's <Electrons> move at $-v$, so $I_2=-I_1$. The positive charges remain lab-stationary and still feel zero external electrostatic <force> in the lab. The appropriate rest frame for wire 2's <Electrons> now moves at $-v$.
In this frame, wire 1's total line charge is
$$
\lambda_1''=\gamma\left(0-\frac{(-v)I_1}{c^2}\right)
=-\gamma\lambda\frac{v^2}{c^2}<0.
$$
It produces an <electric field> pointing toward wire 1. The stationary test <Electron> is negatively charged, so its electric <force> points away from wire 1. Transforming back again divides the individual transverse <force> by $\gamma$, and multiplying by $\lambda/e$ gives
$$
\boxed{\frac{F_x}\ell=+\frac{\lambda^2v^2}{2\pi\epsilon_0c^2d}
=+\frac{\mu_0I^2}{2\pi d},\qquad\text{repulsive}.}
$$
The changed sign arises from the changed transformed charge imbalance, without needing a magnetic-force calculation. More generally the signed result is $F_x/\ell=-\mu_0I_1I_2/(2\pi d)$ for the chosen outward radial direction.
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