Let $A$ be an associative algebra over $\mathbb{C}$ with irreducible finite-dimensional representations on $V$ and $W$. Then is the tensor product of representations on $V \otimes W$ semi-simple?
The below question is concerning representations of a group and in this case the answer is positive due to a result of Chevalley. I tried looking in the literature but I could not find any equivalent results for representations of an algebra.
By a famous theorem of R. Steinberg, STEINBERG, R. Complete sets of representations of algebras. PROC. Am. Math. Soc. 13 (1962), 746-747, if $V$ is a faithful representation for a monoid $M$, then $T(V)=\bigoplus_{n\geq 0}V^{\otimes n}$ is a faithful representation for the monoid algebra $KM$ for any field $K$. If $M$ is finite, then $KM$ is finite dimensional and so it follows easily that $\bigoplus_{n=0}^rV^{\otimes n}$ is a faithful $KM$-module for some $r$.
So you can get a counterexample by taking a finite monoid $M$ with a faithful complex semisimple representation which does not have a semisimple complex algebra. There are many such examples. Here is an easy one.
Consider $M$ the monoid of $3\times 3$ complex matrices consisting of zero, the identity, all matrix units $E_{ij}$ and the nilpotent matrix $\begin{bmatrix}0 & 1 &0\\ 0 &0 &1 \\ 0 & 0 &0 \end{bmatrix}$. Then $\mathbb CM$ has a faithful irreducible representation $V$, namely the one I used to define it. Irreducibility follows since all the matrix units are in the image of the representation, and so the map $\mathbb CM\to M_3(\mathbb C)$ is sujective.
But $\mathbb CM$ does not have a semisimple algebra. If you factor out the ideal of $\mathbb CM$ spanned by the matrices of rank at most 1, you get an algebra with a nilpotent ideal (spanned by the coset of the nilpotent matrix). So $\mathbb CM$ has a nonsemisimple quotient and hence is not semisimple. It follows that not all tensor powers of $V$ are semisimple by R. Steinberg's theorem.
Added. I think already $V\otimes V$ is not semisimple. It contains as a submodule the exterior power $\Lambda^2(V)$. This is annihilated by all the rank $\leq 1$ matrices but not by the nilpotent matrix. So the image of $\mathbb CM$ under this exterior power representation is not semisimple (it has a nilpotent ideal spanned by the image of the nilpotent matrix) and so this subrepresentation is not semisimple and hence $V\otimes V$ is not semisimple.
Added. Here is a two-dmensional example. Take the monoid consting of the 2x2 identity matrix and the matrices $\pm \begin{bmatrix} 1 & 1\cr 0 & 0\end{bmatrix}, \pm \begin{bmatrix} 1 & -1\cr 0 & 0\end{bmatrix}, \pm \begin{bmatrix} 0 & 0\cr 1 & 1\end{bmatrix}, \pm \begin{bmatrix} 0 & 0\cr 1 & -1\end{bmatrix}$. It generated as a monoid by $\begin{bmatrix} -1 & -1\cr 0 & 0\end{bmatrix}, \begin{bmatrix} 0 & 0\cr 1 & -1\end{bmatrix}$. It is easy too see there is no invariant subspace since these rank $1$ matrices span all $2\times 2$ matrices. So this is an irreducible represention $V$.
But $V\otimes V$ has a nonsemisimple submodule. The vectors $e_1\otimes e_1$ and $e_2\otimes e_2$ span an invariant subspace with $e_1\otimes e_1-e_2\otimes e_2$ spanning an invariant subspace $W$ with no complement. A complement would have to be fixed by all of the matrices (since $W/\mathbb C(e_1\otimes e_1-e_2\otimes e_2)$ is the trivial module) but no such vector exists.
Here is an example in which $A$ is freely generated by two elements $x,y$. Let $W$ be $\mathbb C^2$ with representation given by $$ x(a_1,a_2)=(0,a_1) $$$$ y(a_1,a_2)=(a_2,0). $$ This is irreducible. Let $V$ be again two-dimensional, with $x$ and $y$ acting by invertible maps $S:V\to V$, $T:V\to V$ respectively. So $V\otimes W$ can be described as $V\times V$ with $$ x(v_1,v_2)=(0,Sv_1) $$$$ y(v_1,v_2)=(Tv_2,0). $$ A proper nontrivial subrepresentation $U\subset V\otimes W$ corresponds precisely to a one-dimensional subspace $L\subset V$ such that $S(L)=T(L)$; take $U$ to consist of pairs $(v_1,v_2)$ such that $v_1\in L$ and $v_2\in S(L)$.
If $S$ and $T$ are chosen in such a way that $T^{-1}S$ has exactly one $1$-dimensional invariant subspace $L$, then $V\otimes W$ has exactly one proper nontrivial subrepresentation, so it is not semisimple. On the other hand, if we also choose $S$ in such a way that $L$ is not invariant under $S$, then $V$ will be irreducible.
