I asked Dan Freed, who gave a very clean general solution to this problem (as expected).
Here it is (all mistakes in the transcription are mine of course).
The claim is that the equivariant cohomology of G acting on G is indeed the tensor product of cohomology of BG with cohomology of G - in other words the Leray spectral sequence for the fibration $G=\Omega BG \to G/G=LBG \to BG$ degenerates at E_2. To see this we will use the Leray-Hirsch theorem -- i.e. if we can show that every class on the fiber (G) extends to a class on the total space (LBG) then we will be done. Now the cohomology of G is generated (as an exterior algebra) by its primitive classes, and these all come from the generators of the cohomology of BG by transgression. So we just need to show that these transgressed classes actually lift to LBG.
But there is a nice direct construction of these classes on LBG. Namely we use the tautological correspondence $$LBG \leftarrow S^1\times LBG \rightarrow BG$$
where the right arrow is the evaluation map. Thus given a class on BG we can lift it to $S^1 \times LBG$ and then integrate along the circle to get a class on $LBG$. When we restrict these classes to a fiber, i.e. $G=\Omega BG$, we recover the usual transgression construction. (This can be seen very explicitly with differential forms.. the transgression involves the path fibration $$\Omega BG \to P(BG) \to BG$$
and is given by the same kind of tautological/evaluation construction for the map from the interval times the path space to BG, integrating over an interval.. when restricted to $\Omega BG$, ie closed paths, this integration becomes the integration over the circle we had above.)
So we've explicitly lifted all the generators of the cohomology of G to equivariant classes, ie to G/G, hence we're done. (Presumably this can also be seen concretely in the Cartan model, that the generators of H^*G lift to conjugation-equivariant classes..)