= Solution
Reflect the source across the equatorial plane and subtract the two ball kernels. This gives the <Dirichlet Green function inside a hemisphere> $G_{\rm hemi}(\mathbf x;\mathbf x_0)=G_B(\mathbf x;\mathbf x_0)-G_B(\mathbf x;\mathbf x_0^-)$. It is zero on the spherical boundary because each ball kernel is zero there, and on the plane because reflection makes their values equal.
For a <point charge> $e$, <Poisson's equation> in SI units is $\Delta V=-e\delta(\mathbf x-\mathbf x_0)/\varepsilon_0$, so $V=-eG_{\rm hemi}/\varepsilon_0$. Writing $\mathbf x_0^{*-}$ for the reflected inverse point gives
$$
\boxed{V(\mathbf x)=\frac{e}{4\pi\varepsilon_0}\left[\frac1{|\mathbf x-\mathbf x_0|}-\frac1{|\mathbf x-\mathbf x_0^-|}-\frac{a/r_0}{|\mathbf x-\mathbf x_0^*|}+\frac{a/r_0}{|\mathbf x-\mathbf x_0^{*-}|}\right].}
$$
The construction uses the real charge, its negative reflected image, a negative inversion image of magnitude $ea/r_0$, and its positive reflected image. Reflection and inversion commute. All three images lie outside the open upper half-ball, and the prescribed zero boundary values determine the <electrostatic potential> uniquely.
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