I'm not sure exactly what kind of answer you're looking for, but I can try to give some context which may make things sound more reasonable. Let us think of of groupoids as $1$-truncated $\infty$-groupoids, or $1$-truncated spaces. Consider first the case where $A$ is abelian. In this case $BA$ will itself be an $\mathbb{E}_\infty$-group object in spaces, and will have its own classifying space $B^2A$ (which is no longer $1$-truncated, it's a $K(A,2)$). Since the formation of classifying spaces naturally lands in pointed spaces it follows that an arbitrary action of $G$ on $BA$ yields a base-point preserving action of $G$ on $B^2A$. Such a data can be geometrically realized as a fibration $p:F \to BG$ together with an identification $p^{-1}(x_0) \simeq B^2A$ of the fiber over the base point with $B^2A$ and equipped with a $0$-section $s_0:BG \to F$. Furthermore, $F$ will actually form an $\mathbb{E}_\infty$-group object in spaces over $BG$. We note that while we constructed $F$ out of the action of $G$ on $BA$, it actually only depends on the induced action of $G$ on $A$ (something that is well-defined when $A$ is abelian): indeed, as an $\mathbb{E}_\infty$-group object over $BG$, $F$ is just the double delooping of the local coefficients system $BG \to \mathcal{Ab}$ corresponding to the action of $G$ on $A$. The space of sections $\mathrm{Map}_{BG}(BG,F)$ of $F$ is then the space which "represents" the first three cohomology groups of $G$ with coefficients in $A$: namely, $\pi_i\mathrm{Map}_{BG}(BG,F) \cong H^{2-i}(G,A)$ for $i=0,1,2$.

We now turn to the fibration $E \to BG$ that you mention, which is associated to the action of $G$ on $BA$. What is the relation between $E$ and $F$? if the action of $G$ on $BA$ had a homotopy fixed point then $E \to BG$ would have a section and we could think of $F$ as a fiberwise delooping of $E$, or, alternatively, identify $E$ with the fiberwise looping of $F$, i.e., with the homotopy fiber product $BG \times^h_F BG$. In the general case (where we don't assume that there is a base point) one can show that a similar thing happens, only twistedly. Instead of taking the homotopy fiber product of $BG$ with itself over $F$, where the map from $BG$ to $F$ is the $0$-section in both cases, one can show that there actually exists (an essentially unique) other section $s: BG \to F$ such that $E$ is equivalent to the homotopy fiber product $BG \times^h_{F} BG$, where now one copy of $BG$ maps to $F$ via $s$ and the other copy via the $0$-section $s_0$. Roughly speaking, we may consider $E$ as the space of pairs $(x,\eta)$ where $x$ is a point of $BG$ and $\eta$ is path in the fiber $F_x$ from $s(x)$ to the base point $s_0(x) \in F_x$. The section $s$ then determines an element in $\pi_0(\mathrm{Map}_{BG}(BG,F)) \cong H^2(G,A)$, which vanishes if and only if $E \to BG$ admits a section (i.e., if and only if $BA$ has a homotopy fixed point). A choice of such a section is essentially the same as an identification of $E \to BG$ with the fiberwise looping of $F \to BG$, in which case the full space of sections satisfies $\pi_i(\mathrm{Map}_{BG}(BG,E)) \cong H^{1-i}(G,A)$ for $i=0,1$.

This story is quite general, it will work essentially the same if we replace $A$ by any $\mathbb{E}_{\infty}$-group object, or even a spectrum, equipped with an action of $G$ (the cohomology of $G$ with coefficients in such a thing is exactly what you think it should be). Of course, everything works much less well when $A$ is not abelian. First of all, you might ask yourself what do we even mean by $H^i(G,A)$ when $A$ is not abelian. I have to say I don't really have a good answer for this. There exists an explicit definition using cocycles, which for $i=0,1$ requires an honest action of $G$ on $A$ and for $i=2$ only requires an outer action of $G$, namely, a homomorphism $G \to Out(A)$. In this definition $H^0(G,A)$ is a group (the fixed subgroup of $A$ under the action of $G$), $H^1(G,A)$ is a pointed set and $H^2(G,A)$ is a set with a collection of "neutral elements". This definition enjoys some basic properties: it reduces to the usual definition when $A$ is abelian, and admits some useful (not-so-long) long exact sequences. One can also show that, like in the abelian case, the elements of $H^2(G,A)$ are in one-to-one correspondence with extensions of $G$ by $A$ which are compatible with the given outer action, where the neutral elements correspond to the split extensions (note that there can be several split extensions of $G$ by $A$ with the same outer action). In particular, if we are given an action of $G$ on $BA$ then the corresponding fibration $E \to BG$ determines a class in $H^2(G,A)$ via the group extension $ 1 \to A \to \pi_1(E) \to G \to 1$. The corresponding element is neutral if and only if this sequence splits, which is equivalent to $E \to BG$ having a section. A choice of such a section/splitting determines in particular a lift of the outer action of $G$ on $A$ to an honest action, and hence determines a (non-abelian) local coefficient system $BG \to \mathcal{Grp}$. One may then identify $E \to BG$ with the fiberwise delooping of this local coefficient system and show that, as in the abelian case, $\pi_i\mathrm{Map}_{BG}(BG,E) \cong H^{1-i}(G,A)$ for $i=0,1$.

It is worth noting that even though the unabelian $H^2(G,A)$ is not an especially handsome invariant (which makes the obstruction theory above sounds a bit tautological at first), it can occasionally give useful and computable information, and more importantly, it can be shown to vanish under various assumptions. For example, if the map $Aut(A) \to Out(A)$ admits a section and the center $Z(A)$ vanishes (like if $A$ is the alternating group on at least 5 elements) then $H^2(G,A)$ always vanishes. One may then deduce that any action of $G$ on $BA$ admits a homotopy fixed point, a statement that is far from obvious at first sight.