I'm pretty confident the answer to this question is yes, but I am struggling to find a proof.
The definition I'm using is: a $k$-algebra $A$ is separable if for any field extension $K\supseteq k$ we have that $A\otimes_kK$ is semisimple.
I am able to prove that if $A$ and $B$ are both separable, then $A\otimes_k B$ is also separable, but I get stuck at the converse.
For one of my attempts, I took a field extension $K$. I know that $(A\otimes_k B)\otimes_k K$ is semisimple and that $A\otimes_k K$ is semisimple. I tried using the fact that $(A\otimes_k B)\otimes_k K =(A\otimes_k K)\otimes_K (B\otimes_k K)$ to argue that $B\otimes_k K$ must be semisimple, but I then realized that it is possible to have a semisimple algebra and a non-semisimple algebra such that their tensor product is semisimple. So this attempt fails.
I also tried using another characterization of separable $k$-algebras: a $k$-algebra $A$ is separable if $Z(A)\otimes_k \overline{k}\cong \overline{k}\times \ldots \times \overline{k}$, where $\overline{k}$ is an algebraic closure of $k$. I know that $Z(A\otimes_k B)=Z(A)\otimes_k Z(B)$. I also know that $Z(A\otimes_k B)\otimes_k \overline{k} \cong \overline{k}\times \ldots \times \overline{k}$ and that $Z(A)\otimes_k \overline{k} \cong \overline{k}\times \ldots \times \overline{k}$. I somehow want to use this to show that $Z(B)\otimes_k \overline{k} \cong \overline{k}\times \ldots \times \overline{k}$ but I'm not sure how to proceed.
I appreciate any hints or advice. Thanks!
It seems that you are assuming $A$ and $B$ are finite dimensional over $k$.
Suppose first that $B$ is a field. Then $A\otimes_kB$ separable implies $A\otimes_k(B\otimes_kB)$ semisimple. If $B/k$ is not separable, then $B\otimes_kB$ has a non-zero nilpotent ideal, and so the same holds for $A\otimes_k(B\otimes_kB)$, a contradiction. Therefore $B/k$ is separable.
If $B$ is a division algebra with centre $K$, then $A\otimes_kB$ is separable if and only if $A\otimes_kK$ is separable, which by the above is if and only if $K/k$ is separable, if and only if $B/k$ is separable.
In general, $B$ is a finite product of matrix rings over division rings, and again $A\otimes_kB$ is separable if and only if each division ring is separable over $k$, which is if and only if $B/k$ is separable.