= Solution
Write $r=|\mathbf r|$, $s=|\mathbf r'|$. For a spherical source, the angular integral needed for the <Green function of the Laplacian> is
$$
\int\frac{d\Omega'}{|\mathbf r-\mathbf r'|}=2\pi\int_{-1}^1\frac{du}{\sqrt{r^2+s^2-2rsu}}=\frac{4\pi}{\max(r,s)}.
$$
This is the angular form of the <spherical shell theorem>. Substituting the source into the <Green function> integral therefore yields
$$
\phi(r)=-\frac1r\int_0^rs e^{-\beta s}\,ds-\int_r^\infty e^{-\beta s}\,ds.
$$
The elementary integrals are
$$
\int_0^rs e^{-\beta s}\,ds=\frac{1-(1+\beta r)e^{-\beta r}}{\beta^2},\qquad\int_r^\infty e^{-\beta s}\,ds=\frac{e^{-\beta r}}\beta.
$$
The terms proportional to $e^{-\beta r}/\beta$ cancel, leaving
$$
\boxed{\phi(r)=\frac{e^{-\beta r}-1}{\beta^2r},\qquad\phi(0)=-\frac1\beta.}
$$
The value at the origin is the continuous limit. At large radius $\phi\sim-1/(\beta^2r)$, consistent with the integrated source $\int\rho\,dV=-4\pi/\beta^2$ and the factor $1/(4\pi r)$ in the far-field potential.
For a direct check, set $u=r\phi$. Then $u''=e^{-\beta r}$, so $\nabla^2\phi=u''/r=e^{-\beta r}/r=-\rho$ for $r>0$. Boundedness of $\phi$ at zero cancels the singular $1/r$ terms and excludes an extra point source. The original density is itself singular but locally <integrable>; the equation at the origin is understood in the sense of <distributions>. This is the <Poisson potential of an exponential-over-radius source> with amplitude $A=-1$.
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