A variant of #POSITIVE-2-DNF
Note: this reduction is written in the wrong direction, and when fixed it only works for multigraphs. See explanations in the edit to the original question.

I think the problem is #P-hard already on 3-regular graphs using the results of Cai, Lu and Xia, Holographic Reduction, Interpolation and Hardness, 2012. I will do this by showing the #P-hardness of counting the non-satisfying valuations of $G$, i.e., the valuations $\nu$ where for every edge $e$ at least one of the endpoints of $e$ is not mapped to $e$ by $\nu$. Indeed, counting this reduces in PTIME to counting the satisfying valuations of $G$ as you ask: this uses the fact that the total number of valuations (both satisfying and non-satisfying) can be computed in PTIME, using the closed-form formula in your question.

To show the hardness of counting non-satisfying valuations on 3-regular graphs, consider a 3-regular graph $G = (V, E)$, and construct the bipartite graph $G' = (V \cup E, W)$ between $V$ and $E$: it is a 2-3-regular graph in the sense that vertices in $V$ all have degree $3$ and vertices in $E$ all have degree $2$. Now, a non-satisfying valuation of $G$ in your sense amounts to picking one edge of $W$ incident to each vertex of $V$ in $G'$, so that we never pick the two edges of $W$ incident to a vertex of $E$. In other words, I'm claiming that counting the non-satisfying valuations of $G$ is exactly to counting the subsets $W'$ of $W$ such that each vertex of $V$ has exactly one incident edge in $W'$ (= we pick one edge for each vertex of $V$), and each vertex of $E$ has 0 or 1 incident edges in $W'$ (= no edge of $E$ has both its endpoints selected).

If I'm not mistaken, this is precisely the problem #[1,1,0][0,1,0,0] in the notation of Valiant used in the paper I quote: note that there's a hopefully legible explanation in Appendix D of this paper (which, incidentally, we co-authored ;-P). Now looking at the table on page 23 of Cai, Lu and Xia, we see that #[1,1,0][0,1,0,0] is #P-hard.

As for the problem having an established name more palatable than #[1,1,0][0,1,0,0], I don't know, but maybe this can be one direction in which to look.