As you observed, by Lax-Milgram it is easy to show well-posedenss in $H^1(d\mu)$ with a right-side in $H^{-1}(d\mu)$. The constant in the estimate should be given by whatever the optimal constant in the Poincar\'e inequality for the measure is. Now let's try to differentiate the equation. First, we can write $\nabla^* = -\nabla + \nabla V$ and thus
$$
\nabla^*\nabla u = -\Delta u + \nabla V\cdot \nabla u.
$$
Now: formally differentiate the equation
$$
-\Delta u + \nabla V\cdot \nabla u = f
$$
with respect to $x_i$ and put $v := \partial_{x_i}u$:
\begin{equation}
-\Delta v + \nabla V\cdot \nabla v = \partial_{x_i} f-\nabla (\partial_{x_i} V)\cdot \nabla u.
\end{equation}
We now should ask ourselves whether the right side belongs to $H^{-1}(d\mu)$. The answer is yes. First, the term $\nabla (\partial_{x_i} V)$ is bounded because of assumptions on $V$, and the $\nabla u$ term belongs to $L^2(d\mu)$, so their product belongs to $L^2(d\mu)$ and hence $H^{-1}(d\mu)$. Now we have to establish that $\partial_{x_i}f$ belongs to $H^{-1}(d\mu)$. So pick a test function $\phi\in H^1(d\mu)$ with unit norm and test:
$$
\left| \int \partial_{x_i} f \phi \,d\mu \right|
=
\left| \int f \left( \partial_{x_i}\phi - \phi \partial_{x_i} V \right) \,d\mu \right|.
$$
Obviously the term $f\partial_{x_i}\phi$ is ok by the assumption that $\phi\in H^1(d\mu)$ and $f\in L^2(d\mu)$. The second term is also ok because $\partial_{x_i} V$ grows at most like $C(|x|+1)$ and $(1+|x|)\phi(x)$ belongs to $L^2(d\mu)$ by the logarithmic Sobolev inequality.

This is so far purely formal: the argument only works as an a priori estimate. I do not believe it is a good idea to use difference quotients in making this rigorous, since these don't play well with the spaces due of the Gaussian-like decay of the measure. I think you can work backwards: Since we have established that $\partial_{x_i} f-\nabla (\partial_{x_i} V)\cdot \nabla u$ belongs to $H^{-1}(d\mu)$, solve the equation for it and find a solution $v_i$. Then try to argue that $v_i = \partial_{x_i} u$. (I consider this last step more of a formality so I won't go on, but I can think harder about the right way to justify it if needed.)