First, stress-energy tensor (of matter fields) $T_{\mu \nu}$ is something that you have to put in by hand in Einstein's equations:
$$R_{\mu \nu} - \frac{1}{2} R g_{\mu \nu} = \frac{8\pi G}{c^4} T_{\mu \nu}$$
to see how it determines the curvature $(g_{\mu \nu})$. Charge is not to be treated exclusively from $T_{\mu \nu}$, as you seem to think. Everything that can contribute to $T_{\mu \nu}$ must be included in it. If the stress-energy tensor is zero, it implies that the geometry is Ricci flat: $R_{\mu \nu}=0$. (Note that spacetime can still be curved for $R_{\mu \nu}=0$ because $R_{\mu \nu \rho \sigma} \neq 0$, in general).
Now, a charge creates an electric field around it. The electromagnetic (electric only, for our case) field is described by the Lagrangian for classical electromagnetism: $\mathcal{L} = -\frac{1}{4} F^{\mu \nu} F_{\mu \nu}$. To find the $T_{\mu \nu}$ for this electromagnetic field, we need to vary the action for $\mathcal{L}$ with respect to the metric tensor: $T_{\mu \nu} \sim \frac{\delta S}{\delta g^{\mu \nu}}$. This electromagnetic stress-energy tensor is not zero. (It is traceless, however, so the Ricci scalar $R=0$). So $R_{\mu \nu} \neq 0$ and spacetime is curved.
The typical example given for such spacetimes is the Reissner–Nordström metric, which can be derived from above calculations, and some other assumptions.
