Regular elementary abelian subgroups of primitive permutation groups
This is a bit long for a comment. To answer the question it would be very useful to know what groups can act primitively on a set of order $2^a$. Looking at O'Nan--Scott--Aschbacher, we see that such groups must be either affine or almost simple, or product action. Let's consider these three families one at a time.
Almost simple. In this case, the following paper is useful:
Robert M. Guralnick, MR 700286 Subgroups of prime power index in a simple group, J. Algebra 81 (1983), no. 2, 304--311.
The main theorem of the paper classifies the objects in the title. A very useful corollary for us is this:
Corollary 2. Let $G$ be a nonabelian simple group acting transitively on $\Omega$ with $|\Omega|=p^a$ for some prime $p$. Then $G$ acts $2$-transitively on R unless $G \cong {\rm PSU}_4(2)$ and $|\Omega| = 27$ in which case $G$ acts as a rank $3$ primitive permutation group and $G$ has orbits of size $1$, $10$, and $16$.
This result means that, for the purposes of the question we can forget about simple groups containing regular elementary abelian subgroups -- they will never stop these subgroups being $B$-groups.
Affine. Now, as you point out, this reduces to studying irreducible subgroups $S$ of ${\rm GL}_a(2)$ that are intransitive on non-zero vectors. As Derek says we can assume that $a$ is prime and that $2^a-1$ is prime (otherwise $E_{2^a}$ cannot be a $B$-group).
Now, as Derek says, looking at Aschbacher's classification of subgroups of ${\rm GL}_a(q)$ (or maybe Nori's results would suffice), we can assume that $S$ is almost simple. So we're interested in almost simple groups $S$ that have a representation over $\mathbb{F}_2$ of degree a prime $p$ with $2^p-1$ prime, and are intransitive on non-zero vectors.
We can deal with that last condition -- intransitivity on non-zero vectors -- by referring to the main theorem in Guralnick's paper cited above. If $S$ is transitive on non-zero vectors, then this is a primitive action of prime degree and so is on Guralnick's list. I think with a little work you can probably rule out (nearly) all of the possibilities on this list, i.e. (with some possible exceptions) any almost simple group will be intransitive on vectors.
Product action. In this case the group $G\leq U\wr S_k$ where $U$ is almost simple with an action of degree $2^b$ with $b$ a proper divisor of $a$. In particular $a$ is composite, and we already know that groups $E_{2^a}$ with $a$ composite cannot be $B$-groups.
CONCLUSION. The only outstanding case is the affine one which means that we've reduced Question 1 to Question 2. And, given the progress on the affine case described above, based on Derek's remarks we would conclude the following (modulo the exceptions in the final paragraph of the affine case):
$E_{2^a}$ is a $B$-group if and only if $a$ is prime, $2^a-1$ is prime, and there exists no almost simple group (except ${\rm GL}_a(2)$) with an irreducible representation of degree $a$ over $\mathbb{F}_2$.
The problem of ascertaining when almost simple groups have such representations is currently beyond me.
This post gives a complete proof of the final box in Nick Gill's excellent answer. Let $V = \mathbb{F}_2^n$.$\newcommand{\GL}{\mathrm{GL}}$
Theorem. $C_2^n$ is a B-group if and only if $2^n-1$ is a Mersenne prime and the only simple groups having an $n$-dimensional irreducible representation over $\mathbb{F}_2$ are $C_{2^n-1}$ and $\GL(V)$.
The key results we need are
- A transitive permutation group of prime degree is either solvable (and so affine) or $2$-transitive; this is due to Burnside.
- If $H \le \GL(V)$ is $2$-transitive then either $n=4$ and $H \cong A_7$ or $H = \GL(V)$; this was proved by Cameron and Kantor in 1979.
- An almost simple group of degree $2^n$ is $2$-transitive; this follows from Guralnick's Theorem in Nick's answer.
Proof. Suppose $2^n-1$ is composite. Let $g \in \mathrm{GL}(V)$ be a Singer element of order $2^n-1$. By Zsigmondy's Theorem, there is a power $g^r$ such that $\langle g^r \rangle$ acts irreducibly but intransitively on $V$. (This was noted by Derek Holt in a comment on the question.) Hence $V \rtimes \langle g^r \rangle$ is simply transitive and primitive and so $C_2^n$ is not a B-group.
Conversely, suppose $2^n-1$ is prime and that $G$ is a primitive permutation group of degree $2^n$. By the O'Nan–Scott theorem, $G$ is either almost simple, or of affine type, or of diagonal type, or of twisted wreath or product type. Following Nick's answer, in the final two cases, we have $G \le S_a \wr S_b$ with $a^b = 2^n$. If $a=2^m$ then $n= mb$. But $n$ is prime, so this is impossible. The diagonal case does not arise because $2^n$ is not the order of a simple group. By (3) if $G$ is almost simple then $G$ is $2$-transitive.
We are left with the affine case. Suppose that the simple group $S$ has an irreducible $n$-dimensional representation over $\mathbb{F}_2$. Then $V \rtimes S$ is primitive and simply transitive, so $C_2^n$ is not a B-group.
Finally, suppose that $C_2^n$ is not a B-group. We must find a proper non-cyclic irreducible simple subgroup $S$ of $\GL(V)$. We are in the affine case, so there exists an irreducible $H \le \GL(V)$ such that $H$ is intransitive on $V \backslash \{0\}$. Let $M$ be a maximal subgroup of $\GL(V)$ containing $H$. The 10 irreducible Aschbacher classes all involve groups preserving a decomposition of $V$ that can exist only when $\dim V = n$ is composite. Therefore $M$ is almost simple.
If $M$ is transitive on $V \backslash \{0\}$ then, since $|V \backslash \{0\}| = 2^n-1$ is prime, (1) implies that either $M \cong C_{2^n-1} \rtimes C_s$ for some $s$ dividing $2^n-1$, or $M$ is $2$-transitive. In the first case, since $H$ is irreducible, $H \ge C_{2^n-1}$. In the remaining case $V \rtimes M$ is $3$-transitive, and by (2), $M = \GL(V)$, and so $H = \GL(V)$. In both cases this contradicts the intransitivity of $H$. Hence we did not lose simple transitivity in passing from $H$ to $M$. Let $S$ be the simple subgroup of $M$. By Clifford's Theorem, the restriction of the $\mathbb{F}_2 M$-module $V$ to $S$ is a direct sum of conjugate irreducible submodules. But $\dim V = n$ is prime. Hence $S$ acts irreducibly. Since $M$ is intransitive on $V \backslash \{0\}$, so is $S$. Hence $S$ is as required. $\Box$
Remark. It seems plausible that the only simple groups having an $n$-dimensional irreducible representation over $\mathbb{F}_2$ for $n$ such that $2^n-1$ is a Mersenne prime are $C_{2^n-1}$ and $\GL(V)$. From the modular Atlas data in GAP, the only sporadic simple groups that could be exceptions are $J_4$, $Ly$, $Th$, $Fi_{24}$, $B$ and $M$. Tables of small dimensional representations of quasi-simple groups in non-defining characteristic and Chevalley groups in defining characteristic, show there are no exceptional representations if $n \le 250$. So if $n \le 250$ then $C_2^n$ is a B-group if and only if
$$ n \in \{1,2,3,5,7,13,17,19,31,61,89,107,127\}. $$
Edit. I had to edit this answer to remove the entirely false claim in the final remark that all irreducible representations of alternating groups over $\mathbb{F}_2$ are self-dual (and so have an invariant symplectic form and are even dimensional). The isomorphism $A_8 \cong \mathrm{GL}_4(\mathbb{F}_2)$ shows this is false.