If $A$ over $\mathbb{C}$ is an abelian variety of dimension $g$, then for all $i \in \mathbb{N}$, $H_i(A,\mathbb{Q}) \cong H^i(A,\mathbb{Q}) \cong \mathbb{Q}^{ {2g \choose i}}$.

Taking $i = 1$, the conclusion you are asking about is true if and only if $\operatorname{dim} A_0 \otimes_\mathbb{Q} E = [E:\mathbb{Q}] \operatorname{dim} A_0$.

Which it probably is, if $A_0 \otimes_{\mathbb{Q}} E$ means the Serre tensor construction. I can't quite remember how this goes at the moment (and I'll wait for you to confirm your notation before trying).

If you tell/remind us exactly what $A_0 \otimes_{\mathbb{Q}} E$ means, we could probably give you a natural isomorphism between these two homology groups.

If $A$ over $\mathbb{C}$ is an abelian variety of dimension $g$, then for all $i \in \mathbb{N}$, $H_i(A,\mathbb{Q}) \cong H^i(A,\mathbb{Q}) \cong \mathbb{Q}^{ {2g \choose i}}$.

Taking $i = 1$, the conclusion you are asking about is true if and only if $\operatorname{dim} A_0 \otimes_\mathbb{Q} E = [E:\mathbb{Q}] \operatorname{dim} A_0$.

Which it probably is, if $A_0 \otimes_{\mathbb{Q}} E$ means the Serre tensor construction. I can't quite remember how this goes at the moment (and I'll wait for you to confirm your notation before trying).

If you tell/remind us exactly what $A_0 \otimes_{\mathbb{Q}} E$ means, we could probably give you a natural isomorphism between these two homology groups.

Addendum: I found a nice online treatment of Serre's tensor construction here.