= Solution
In $S$, adding the stationary opposite line cancels <charge density> and <electric field> but contributes no current. Hence
$$
\lambda_{\rm tot}=0,\qquad I_{\rm tot}=\lambda u,\qquad
\mathbf E_{\rm tot}=0,\qquad
\mathbf B_{\rm tot}=\frac{\mu_0\lambda u}{2\pi r}\widehat{\boldsymbol\varphi}.
$$
The <neutral current filament in a boosted frame> is not neutral. Transform its line-integrated <four-current>:
$$
\boxed{\lambda'_{\rm tot}=-\frac{\gamma u I_{\rm tot}}{c^2}=-\gamma\lambda\frac{u^2}{c^2},\qquad
I'_{\rm tot}=\gamma I_{\rm tot}=\gamma\lambda u.}
$$
The same result follows by adding the original resting line $(\lambda',0)$ to the new negative line, whose primed <density> and current are $(-\gamma\lambda,+\gamma\lambda u)$. In particular $\lambda'-\gamma\lambda=-\gamma\lambda u^2/c^2$ because $\lambda=\gamma\lambda'$.
Transforming the total fields gives
$$
\boxed{\mathbf E'_{\rm tot}=-\frac{\gamma\lambda u^2}{2\pi\epsilon_0c^2r}\widehat{\mathbf r},\qquad
\mathbf B'_{\rm tot}=\frac{\mu_0\gamma\lambda u}{2\pi r}\widehat{\boldsymbol\varphi}.}
$$
For positive $\lambda$, $\mathbf u\times\mathbf B$ points inward, explaining the electric-field sign. The primed Gauss and Ampère fields calculated from $\lambda'_{\rm tot}$ and $I'_{\rm tot}$ reproduce these expressions. Both a moving charge distribution and a moving current must be transformed; boosting just one constituent would give an inconsistent total.
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