Let $U$ be sufficiently smooth, $\beta$ a constant and $$ \mathcal{L}p = \frac{1}{\beta}\Delta p + \nabla\cdot(p\nabla U)\\ \mathcal{L}^*g = \frac{1}{\beta}\Delta g - \nabla g \cdot\nabla U. $$ Now I want to show that $\mathcal{L}e^{-U}g = e^{-U}\mathcal{L}^*g$.
Using the Laplacian product rule everything just started to look even worse, i.e.:
$$\mathcal{L}e^{-U}g = \frac{1}{\beta}(e^{-U}\Delta g + 2\nabla e^{-U}\nabla g + \Delta e^{-U} g) + \nabla e^{-U}g\nabla U + e^{-U}\nabla g\nabla U + e^{-U} g \Delta U$$
The $\frac{1}{\beta}e^{-U}\Delta g$ terms would cancel out just fine, but I don't see how to get rid of the other terms with a $\frac{1}{\beta}$.
Factoring out the $e^{-U}$ does not even work in the one dimensional case ($\nabla = \frac{d}{dx}$); there I get:
$$e^{-U}\left(\frac{1}{\beta}(g''-2Ug'+U'^2-U'')-gUU'+g'U'+gU''\right)$$
which doesn't really resemble
$$e^{-U}\left(\frac{1}{\beta}g''-g'U'\right).$$
(I would like to comprehend equation (2.16) in this paper)
I think that there might be a missing $\beta$ in the exponent $e^{-U}$ meaning it could be that the relation that they mean is $$ \mathcal{L}e^{-\beta U}g=e^{-\beta U}\mathcal{L^*}g. $$ At least it seems that their relation holds if one replace $\beta$ by $1$.