Ok, I think one can quickly reduce to the case where $A$ is local and also that $B$ is $A$-free of finite rank (I need to think about this latter assumption a little, but probably by completion we can do this).
But granting these assumptions, we, can proceed as follows:
Set $A \to A'$ to be the absolute Frobenius on $A$ and likewise $B \to B'$ the Frobenius on $B$. The primes will just help me distinguish source and target.
Then we want to show that the obvious map: $$A' \otimes_A B \to B'$$ is an isomorphism.
Now, $B'$ is a free $A'$ module, likewise so is $A' \otimes_A B$. So we are trying to establish that a certain map of free $A'$-modules is an isomorphism. First we show surjectivity, so we mod out by the maximal ideal $m' \subseteq A'$. Then we obtain: $$\Phi : A'/m' \otimes_{A/m} B/m \cong A'/m' \otimes_A B \to B'/m'B'.$$ Now, $A/m$ is a field, and $B/mB$ is a separable extension (by the etale hypothesis). $A'/m'$ is a purely inseparable extension and so we easily see (by using that the extensions are linearly disjoint) that $B'/m'B'$ is identified naturally with $A'/m' \otimes_{A/m} B/m$.
It follows that $\Phi$ is an isomorphism. Thus $A' \otimes_A B \to B$ is surjective by Nakayama's Lemma. But $A' \otimes_A B$ and $B'$ have the same rank as $A'$-modules by the above argument. But then we have a surjection between finite free modules of the same rank. Thus the map is an isomorphism.