Yes. The principal bundles are the same and your guess that $BA$ is an abelian group is exactly right. A good reference for this story, and of Segal's result that David Roberts quotes, is Segal's paper:
G. Segal. Cohomology of topological groups, Symposia Mathematica IV (1970), 377- 387.
The functors $E$ and $B$ can be described in two steps. First you form a simplicial topological space, and then you realize this space. It is easy to see directly that $EG$ is always a group and that there is an inclusion $G \rightarrow EG$, which induces the action. The quotient is $BG$. Under suitable conditions, for example if $G$ is locally contractible (which includes the discrete case), the map $EG\rightarrow BG$ will admit local sections and so $EG$ will be a $G$-principal bundle over $BG$. This is proven in the appendix of Segal's paper, above. There are other conditions (well pointedness) which will do a similar thing.
The inclusion of $G$ into $EG$ is a normal subgroup precisely when $G$ is abelian, and so in this case $BG$ is again an abelian group.
I believe your question was implicitly in the discrete setting, but the non-discrete setting is relevant and is the subject of Segal's paper. Roughly here is the answer: Given an abelian (topological) group $H$, the $BH$-princical bundles over a space $X$ are classified by the homotopy classes of maps $[X, BBH]$. When $H$ is discrete, $BBH = K(H,2)$. If $X = K(G,1)$ for a discrete group $G$, these correspond to (central) group extensions:
$$H \rightarrow E \rightarrow G$$
If $G$ has topology, then the group extensions can be more interesting. For example there can be non-trivial group extensions which are trivial as principal bundles. Easy example exist when H is a contractible group. However Segal developed a cohomology theory which classifies all these extensions. That is the subject of his paper.
The argument in the quoted paper is a bit too sketchy. An actual proof will have two parts:
Let $f:X \to Y$ be map. Assume that for each $y \in Y$, $\tilde{H}^{\ast} (f^{-1}(y);\mathbb{Q})=0$ (this is what the phrase ''$\mathbb{Q}$-acyclic'' means). Find conditions that guarantee that $f$ induces an isomorphism in cohomology. Essentially, one needs to provide assumptions that the Leray spectral sequence of the map is as nice as claimed in Marks answer.
Provide conditions for group action of a topological group $G$ on a space $X$ such that $EG \times_G X \to X/G$ has the properties found in 1 (and the answer in 1 depends on the example you want to consider).
ad 1. Just having finite stabilizers is not enough. Consider the action of $G=\mathbb{Z}$ on $X=S^1$ by an irrational rotation. Then $EG \times_G S^1 \simeq S^1 \times S^1$. The quotient space is an uncountable set with the trivial topology. Because each map into this space is continuous, it is contractible and has trivial (Cech or singular) cohomology.
One needs criteria that show that a map $f:X \to Y$ with contractible/acyclic/$\mathbb{Q}$-acyclic fibres is itself a weak equivalence/acyclic/$\mathbb{Q}$-acyclic. Having contractible fibres does not suffice: look at the identity map $[0,1]^{\delta} \to [0,1]$, where $\delta$ means ''discrete topology''.
Such problems are solved by what I would call ''fibre theorems''. You wish to know how close the natural map $f^{-1}(y) \to hofib_f (y)$ from the point-preimages to the homotopy fibres is to be a weak homotopy equivalence. What one needs is a suitable local condition and then a local-to-global result. Many technical results are of this type: ''fibre bundles are Serre-fibrations'', ''local Hurewicz fibrations are Hurewicz fibrations'', Dold-Thom's result on quasifibrations, the Vietoris-begle mapping theorem, Quillens Theorems A and B and McDuff-Segals formulation of the ''group-completion'' theorem. A variation of the latter solves your problem:
Assume, as above, that for each $y \in Y$, $f^{-1}(y)$ is $\mathbb{Q}$-acyclic.
Assume moreover that $Y$ is Hausdorff and for each $y \in U \subset Y$, there is $y \in V \subset U$ ($U$ and $V$ open) such that $V$ is contractible and $ f^{-1} (V)$ is $\mathbb{Q}$-acyclic (let us call these sets ''good''). Then, I claim, $f$ induces an isomorphism in rational cohomology.
Proof: Let $P:=\coprod_U U$, where $U$ runs through all good sets. This is a topological poset: $(U,x) \leq (V,y)$ iff $x=y$ and $U \subset V$. Take the nerve $|N_{\bullet} P|$ and consider the map $g:|N_{\bullet} P| \to Y$. It has contractible point-preimages: the preimage of $x$ is the poset of all good sets containing $x$, ordered by inclusion and the assumption says that this has contractible realization. In their recent paper arXiv:1201.3527, Galatius and Randal-Williams prove a lemma that under the present assumptions, $g$ is a Serre fibration with contractible fibres, in particular a weak homotopy equivalence.
Apply the same argument to get a poset $Q$, formed not from good subsets of $Y$, but their preimages in $X$. Thus up to weak equivalence, we can replace $f:X \to Y$ by $|N_{\bullet} Q| \to |N_{\bullet} P|$. By assumption, on the $p$th simplicial stage, the map $N_p Q \to N_p P$ is a rational homology equivalence (because this is just a disjoint union of the maps $f^{-1}(U) \to U$). The spectral sequence of simplicial spaces gives then that $|N_{\bullet} Q| \to |N_{\bullet} P|$ is a rational homology equivalence.
ad 2. There are several instances where the above argument can be applied. Example 1: let $G$ be a discrete group that acts simplicially on a simplicial complex, with finite stabilizers.
Example 2: let $G$ be a Lie group that acts properly on a smooth manifold with finite stabilizers. By some standard results, there is a $G$-invariant Riemannian metric and $G$-equivariant tubular neighborhoods of orbits $Gx\cong G/G_x$. The image of such a tubular neighborhood $U \subset M$ is a contractible subset of $M/G$. The preimage of $U/G$ under $EG \times_G M \to M/G$ is $EG \times_G U \simeq EG \times_G G/G_x \simeq BG_x$, which is $\mathbb{Q}$-acyclic.
Best Answer
Added Aug 2016: I've written this up, available at https://arxiv.org/abs/1608.02999
$\def\Hom{\mathrm{Hom}} \def\Map{\mathrm{Map}} \def\ad{\mathrm{ad}}$
I think this is true. I'll sketch a possible proof here. I haven't carefully checked everything, and there are things that need checking. Feel free to do that.
First, we can assume $H=G$: we want to show that $(B_G\Pi)^G\to \mathrm{Map}(BG,B\Pi)$ is an equivalence if $G$ is compact Lie and $\Pi$ is compact Lie and a 1-type.
We might as well consider the induced map on homotopy fibers over the maps to $B\Pi$ (induced by evaluating at the basepoint of $BG$.) That is, we want to show $$ \Hom(G,\Pi) \to \Map_*(BG,B\Pi) $$ is a weak equivalence. Here $\Hom(G,\Pi)$ is topologized as a subspace of $\Map(G,\Pi)$. We know that this is homeomorphic to $\coprod_{[\phi]} \Pi/C_\Pi(\phi(G))$, a coproduct over conjugacy classes of homomorphisms $G\to \Pi$ (see Nearby homomorphisms from compact Lie groups are conjugate).
Given this, it is already clear we get an equivalence when $G$ is compact and connected (reduce to the case where $\Pi$ is a torus). It's not so easy to see why this is so for general $G$: although we can "compute" both sides, the accounting is different and hard to match up.
Here's an attempt at a general proof, based on the ideas which work when $\Pi$ is a torus (which involve the idea of continuous cochains as in Graeme Segal, "Cohomology of topological groups"). It should fit into some already-known technology (cohomology of topological groups with coefficients in a topological 2-group?), but I don't want to bother to figure out what or how.
Consider data consisting of
a group $\Pi$ (a compact Lie 1-type as above), with connected component $\Pi_0$ (which is abelian),
a vector space $V$,
(It's a kind of crossed module.) Given this, define $E(G, (\Pi,V))$ to be the space of pairs $(f,v)$ where $f\colon G\to \Pi$ and $v\colon G\times G\to V$ are continuous maps, satisfying
(I might want to additionally require a normalization: $f(e)=e$. Or not.)
The examples I have in mind are $E:= E(G, (\Pi, T_e\Pi))$ and $E^0:= E(G, (\Pi,0))$. The claims are as follows.
$E$ is weakly equivalent to $\Map_*(BG,B\Pi)$. To compute the mapping space, you need to climb the cosimplicial space $[k] \mapsto \Map_*(G^k, B\Pi)$. Because $B\Pi$ is a 2-type, you don't need to climb very far. The idea is that if you do this, and you keep in mind facts such as:
$\Pi$ is equivalent to $\Omega B\Pi$, and
the fibration $(v,\pi)\mapsto (\pi, \exp[v]\pi) \colon V\times\Pi\to \Pi\times \Pi$ is equivalent to the free path fibration $\Map([0,1],\Pi)\to \Pi\times \Pi$,
you see that you get an equivalence. (I came up with the definition of $E$ exactly by doing this.)
That's kind of sketchy.
More concretely: $\Map_*(BG,B\Pi)$ can be identified with the space of maps between pointed simplicial spaces, from $G^\bullet$ to $S_\bullet:=\bigl([n]\mapsto \Map_*(\Delta^n/\mathrm{Sk}_0\Delta^n, B\Pi)\bigr)$. The space $E$ is also a space of maps between such, from $G^\bullet$ to $N_\bullet$, where $N_\bullet$ is a simplicial space built from the crossed module $(\Pi,V)$ (the nerve of the crossed module, as in https://mathoverflow.net/q/86486 ) with $N_n \approx \Pi^n\times V^{\binom{n}{2}}$. It's not to hard to show that $N_\bullet$ and $S_\bullet$ are weakly equivalent Reedy fibrant simplicial spaces; they both receive a map from $\Pi^\bullet$, which exhibits the equivalence. (But note: showing that $N_\bullet$ is Reedy fibrant relies crucially on the fact that $\exp$ is a covering map.)
$E^0$ is homeomorphic to $\Hom(G,\Pi)$. Yup.
The inclusion $E^0\subseteq E$ is a weak equivalence.
To see this, let $C^1:=\Map(G,V)$, as a topological group under pointwise addition. There is an action $C^1\curvearrowright E$, by $u\cdot (f,v)=(f',v')$ where
It's useful to note that for any $(f,v)\in E$, the resulting map $G\xrightarrow{f} \Pi\to \Pi/\Pi_0$ is a homomorphism. Thus we write $E=\coprod E_\gamma$ for $\gamma\in \Hom(G,\Pi/\Pi_0)$, and $C^1$ acts on each $E_\gamma$.
Consider $(f,0)\in E_\gamma^0= E_\gamma\cap E^0$. Note that $u\cdot (f,0)$ has the form $(f',0)$ for some $f'$ if and only if $u\in Z^1_\gamma$, where this is the set of $u\colon G\to V$ such that
So the action passes to an injective map $C^1\times_{Z^1_\gamma} E^0_\gamma\to E_\gamma$. In fact, it should be a homeomorphism. To see that it's surjective, fix $(f,v)\in E_\gamma$; we need to solve for $u\in C^1$ such that
This amounts to the vanishing of $H^2$ in the complex $C^\bullet_\gamma$ of continuous cochains: $C^t_\gamma:=\Map(G^{t}, V_\gamma)$ (where the differential uses the action $\ad\gamma\colon G\to\mathrm{Aut}(V)$). The vanishing is because $G$ is compact, so we can "average" over Haar measure to turn a non-equivariant contracting homotopy on $D^\bullet_\gamma=\Map(G^{\bullet+1}, V_\gamma)$ into a contracting homotopy on $C^\bullet_\gamma = (D^\bullet_\gamma)^G$.
Given this, since both $C^1$ and $Z^1_\gamma$ are contractible groups, (in fact, $Z^1_\gamma=V/V^{\gamma(G)}$ by $H^1=0$), we should have that $C^1\times_{Z^1_\gamma} E^0_\gamma$ is weakly equivalent to $E^0_\gamma$.
Note: in the case that $\Pi$ is abelian, we simply get a homeomorphism $C^1\times E^0\approx E$.