The answer is yes, these constructions make sense also for stacks, although you typically obtain a power series rather than a polynomial. Funnily enough, the best reference I know is Deligne's Hodge III, even though it doesn't talk at all about algebraic stacks.

Before asking how to put a mixed Hodge structure on the Betti cohomology of a complex algebraic stack $\mathfrak X$, it makes sense to ask how to define the Betti cohomology in the first place. There are several possible approaches. The most direct is to choose an atlas $U \to \mathfrak X$. This choice gives rise to a simplicial scheme $U_\bullet$, with $U_n = U \times_{\mathfrak X} U \times_{\mathfrak X} \ldots \times_{\mathfrak X} U$ ($n+1$ factors). Taking levelwise analytification gives a simplicial topological space, which can be geometrically realized to a bona fide topological space. If I had chosen a different presentation I would have gotten a weakly equivalent (in fact homotopy equivalent) geometric realization, so the homotopy type of the realization makes sense as an invariant of $\mathfrak X$. We may then take its cohomology.

The point is, then, that Deligne in Hodge III tells us precisely how to put a mixed Hodge structure on the cohomology of a simplicial scheme (of finite type over $\mathbf C$). As you may know, Deligne needs to pass to simplicial schemes in order to define the mixed Hodge structure on a singular variety $X$ (via a "hyperresolution of singularities"), but he notes that he obtains a general definition of mixed Hodge structure on any simplicial scheme, and as an example application he computes the mixed Hodge structure on the classifying stack of a linear algebraic group.

Unfortunately I don't know off hand a reference that connects the dots between what's in Hodge III and generalities on algebraic stacks, though it is all certainly something "known to experts". Perhaps you can look into Noohi's papers on topological stacks.

Regarding your question whether $E([X/G])=E(X)/E(G)$, this is not true in full generality. Think about the case when $X$ is a point and $G$ is finite, in which case $E([X/G])=E(X)=1$, but $E(G) = \vert G\vert$. But if you assume that your group $G$ is connected then you're good. You can prove it by analysing the Leray-Serre spectral sequences for the two fibration sequences
$$ G \to \mathrm{pt} \to BG $$
and
$$ X \to [X/G] \to BG. $$
The first sequence implies $E(G)E(BG)=1$, and the second that $E(X)E(BG)=E([X/G])$. The key point is that if we assume $G$ connected then $BG$ is simply-connected, so the local systems in the spectral sequence trivialize.