= Solution
Factor
$$
f(r)=\left(1-\frac{r_+^2}{r^2}\right)\left(1-\frac{r_-^2}{r^2}\right),
\qquad r_\pm^2=M\pm\sqrt{M^2-Q^2}.
$$
Introduce $v=t+r_*$ with $dr_*/dr=f^{-1}$. Since $dt=dv-dr/f$,
$$
\boxed{ds^2=-f(r)\,dv^2+2\,dv\,dr+r^2d\Omega_3^2},
\qquad d\Omega_3^2=d\theta^2+\sin^2\theta\,d\phi^2+\cos^2\theta\,d\psi^2.
$$
These are the charged analogue of <ingoing Eddington-Finkelstein coordinates>. The $(v,r)$ block has determinant $-1$, so the metric and its inverse are nonsingular at $r=r_+$. For $M\geq Q$, $r_+$ is real and the metric extends analytically through it. The original potential becomes $A=-Q\,dv/r^2+Q\,dr/(r^2f)$; the radial term is locally exact, and a <gauge transformation> gives the horizon-regular potential $A=-Q\,dv/r^2$.
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