Here is an alternate version of @Mare's answer. First recall that a $K$-algebra $A$ is separable if it is semisimple under all base extensions; its enough to check over an algebraic closure of $K$.

Let us write $L\otimes_K A$ as $A^L$ for a $K$-algebra $A$ and field extension $L/K$.

Any semisimple algebra over a perfect field is separable. By Wedderburn-Artin, it suffices to show that if $D$ is a finite dimensional division algebra over $K$, then $D^{\overline K}=\overline{K}\otimes_K D$ is semisimple, where $\overline{K}$ is an algebraic closure of $K$. Let $F$ be the center of $D$; then $F/K$ is a finite field extension and hence separable because $K$ is perfect. Then $D^{\overline K}\cong (\overline K\otimes_K F)\otimes_F D$. But since $F/K$ is separable, basic field theory says $\overline K\otimes_K F\cong \overline K^{[F:K]}$. Therefore, $D^{\overline K}\cong (\overline K\otimes_F D)^{[F:K]}$. But $D$ is central simple over $F$ and so by a basic result in the theory of central simple algebras, $\overline K\otimes_F D\cong M_n(\overline K)$ where $n^2=[D:F]$. Thus $D^{\overline K}$ is semisimple.

Also, if $A$ is a split semisimple $K$-algebra (so isomorphic to a direct product of matrix algebras over a field), then $A$ is separable. In particular, if $\Lambda = KQ/I$ where $Q$ is a quiver and $I$ is an admissible ideal, then $\Lambda/J(\Lambda)\cong K^{|Q_0|}$ and hence is a separable $K$-algebra.

Thus your question really boils down to proving that if $A$ and $B$ are separable $K$-algebras, then so is $A\otimes_K B$. It is enough to show that $(A\otimes_K B)^{\overline K}$ is semisimple where $\overline K$ is an algebraic closure. But $(A\otimes_K B)^{\overline K}\cong A^{\overline K}\otimes_{\overline K}B^{\overline K}$ as they both have the same universal property. The right hand side is a tensor product of direct sums of matrix algebras over $\overline{K}$ and hence is a direct sum of matrix algebras over $\overline{K}$ (as $M_n(\overline K)\otimes_{\overline K} M_m(\overline {K})\cong M_{mn}(\overline K)$) and thus semisimple. Therefore, $A\otimes_K B$ is separable and hence semisimple.