There is a trick that reduces the equation $u_t=Lu$, $L=\Delta-x \nabla$ to the heat equation $u_t-\Delta$. It is genuinely parabolic and gives the parabolic kernel in the whole space, from which the elliptic kernel can be deduced by integrating in time.
If $v_t(t,x)=\Delta v(t,x)-x\nabla v (t,x)$ with $v(0,x)=f(x)$, then $u(t,x)=e^{-2t}v(t, e^tx)$ solves $u_t(t,x)=\Delta u(t,x)-2u(t,x)$ with $u(0,x)=f(x)$.
It does not work in an a ball where probably an expansion in spherical harmonics can give the result for the elliptic case directly.

EDIT. Sorry for the mistake, but the equation for $u$ is wrong. Let us do it in two steps. First put $u(t,x)=v(t, e^t x)$. Then $u_t(t,x)=e^{-2t}\Delta u(t,x)$ with $u(0,x)=f(x)$, which is a simple non-autonomous heat equation. Then, setting $$u(t,x)=w(\frac {1-e^{-2t}}{2}, x)$$ we have $w_t=\Delta w$ with $w(0,x)=f(x)$.

Therefore the final transformation is $v(t,x)=w(\frac {1-e^{-2t}}{2}, e^{-t}x)$.