I don't know if these groups have been studied before, but I can say something about their cohomology rings, at least over $\mathbb{Q}$ and in a stable range. Namely, for $n \gg k$, we have $H^k(E_n;\mathbb{Q}) = \mathbb{Q}$ if $k=0,1$ and $H^k(E_n;\mathbb{Q}) = 0$ for $k \geq 2$. Of course, this is the same as the cohomology of the ordinary braid group as computed by Arnold in

V. I. Arnold, On some topological invariants of algebraic functions, Trudy Moscov. Mat. Obshch. 21 (1970), 27-46 (Russian), English transl. in Trans. Moscow Math. Soc. 21 (1970), 30-52.

Recall that if $H$ is a finite-index normal subgroup of $G$, then $G$ acts on $H^k(H;\mathbb{Q})$ and using the transfer map we have that $H^k(G;\mathbb{Q})$ is equal to the invariants of this action.

For braid groups, the action of $B_n$ on $H^k(PB_n;\mathbb{Q})$ factors through an action of the symmetric group $S_n$, so $H^k(PB_n;\mathbb{Q})$ is a representation of $S_n$ and $H^k(B_n;\mathbb{Q})$ is the trivial subrepresentation $\{\text{$v \in H^k(PB_n;\mathbb{Q})$ $|$ $\sigma v = v$ for all $\sigma \in S_n$}\}$.

Let's now consider $E_n$. In this case, the above argument shows that $H^k(E_n;\mathbb{Q})$ is the subrepresentation $\{\text{$v \in H^k(PB_n;\mathbb{Q})$ $|$ $\sigma v = v$ for all $\sigma \in A_n$}\}$.

Now, representations of finite groups over $\mathbb{Q}$ decompose into direct sums of irreducible representations. The only two irreducible representations of $S_n$ that restrict to the identity on $A_n$ are the trivial representation and the alternating representation. As we said above, the trivial representation corresponds to $H^k(B_n;\mathbb{Q})$, so we conclude that $$H^k(E_n;\mathbb{Q}) = W \oplus H^k(B_n;\mathbb{Q}),$$ where $W \subset H^k(PB_n;\mathbb{Q})$ is the direct sum of all alternating subrepresentations.

The above calculation is thus equivalent to the assertion that the alternating representation does not occur in $H^k(PB_n;\mathbb{Q})$. It's quite hard to decompose $H^k(PB_n;\mathbb{Q})$ into irreducibles; however the paper "Representation Theory and Homological Stability" (see here) by Church and Farb introduces a recipe that they call "representation stability" which describes how the decomposition of $H^k(PB_{n+1};\mathbb{Q})$ into irreducibles can be constructed from the decomposition of $H^k(PB_n;\mathbb{Q})$ into irreducibles, at least for $n$ large. Their results are hard to summarize briefly, but they do imply that the alternating representation does not occur (it is not "stable" in their sense), again at least for $n$ large.

I don't know if these groups have been studied before, but I can say something about their cohomology rings, at least over $\mathbb{Q}$. Namely, we have $H^k(E_n;\mathbb{Q}) = \mathbb{Q}$ if $k=0,1$ and $H^k(E_n;\mathbb{Q}) = 0$ for $k \geq 2$. Of course, this is the same as the cohomology of the ordinary braid group as computed by Arnold in

V. I. Arnold, On some topological invariants of algebraic functions, Trudy Moscov. Mat. Obshch. 21 (1970), 27-46 (Russian), English transl. in Trans. Moscow Math. Soc. 21 (1970), 30-52.

Recall that if $H$ is a finite-index normal subgroup of $G$, then $G$ acts on $H^k(H;\mathbb{Q})$ and using the transfer map we have that $H^k(G;\mathbb{Q})$ is equal to the invariants of this action.

For braid groups, the action of $B_n$ on $H^k(PB_n;\mathbb{Q})$ factors through an action of the symmetric group $S_n$, so $H^k(PB_n;\mathbb{Q})$ is a representation of $S_n$ and $H^k(B_n;\mathbb{Q})$ is the trivial subrepresentation $\{\text{$v \in H^k(PB_n;\mathbb{Q})$ $|$ $\sigma v = v$ for all $\sigma \in S_n$}\}$.

Let's now consider $E_n$. In this case, the above argument shows that $H^k(E_n;\mathbb{Q})$ is the subrepresentation $\{\text{$v \in H^k(PB_n;\mathbb{Q})$ $|$ $\sigma v = v$ for all $\sigma \in A_n$}\}$.

Now, representations of finite groups over $\mathbb{Q}$ decompose into direct sums of irreducible representations. The only two irreducible representations of $S_n$ that restrict to the identity on $A_n$ are the trivial representation and the alternating representation. As we said above, the trivial representation corresponds to $H^k(B_n;\mathbb{Q})$, so we conclude that $$H^k(E_n;\mathbb{Q}) = W \oplus H^k(B_n;\mathbb{Q}),$$ where $W \subset H^k(PB_n;\mathbb{Q})$ is the direct sum of all alternating subrepresentations.

The above calculation is thus equivalent to the assertion that the alternating representation does not occur in $H^k(PB_n;\mathbb{Q})$. This follows from the calculation of $H^k(PB_n;\mathbb{Q})$ as a representation of $S_n$ which was done by in the paper "Coxeter group actions on the complement of hyperplanes and special involutions" by Felder-Velesov; see here.

The above ref to Felder-Velesov was suggested by Vladimir Dotsenko; I originally included the argument below, which only works for $n \gg k$.

It's quite hard to decompose $H^k(PB_n;\mathbb{Q})$ into irreducibles; however the paper "Representation Theory and Homological Stability" (see here) by Church and Farb introduces a recipe that they call "representation stability" which describes how the decomposition of $H^k(PB_{n+1};\mathbb{Q})$ into irreducibles can be constructed from the decomposition of $H^k(PB_n;\mathbb{Q})$ into irreducibles, at least for $n$ large. Their results are hard to summarize briefly, but they do imply that the alternating representation does not occur (it is not "stable" in their sense), again at least for $n$ large.