This is a classic result of Chevalley, and I think the following purely algebraic proof technique is due to Serre. The beautiful trick, which grew into a powerful method, is to take Zariski closures of the images of $G$ in various ${\rm{GL}}_n$'s and to exploit the structure theory of algebraic groups (including the finiteness of component groups thereof) to separately reduce the problem to the easy cases of finite groups and tori and the case of representation theory of semisimple Lie algebras. It is the possibility to bring in Lie algebra considerations for representations of a given "discrete" group that is the great idea.

I prefer to write $\Gamma$ rather than $G$ to denote the given discrete group, so our setup will be a pair of semisimple finite-dimensional representations $(V, \rho)$ and $(V', \rho')$ of $\Gamma$ over a field $k$ of characteristic 0. We want to show that $V \otimes V'$ equipped with the natural representation $r$ defined by the condition $r(\gamma): v \otimes v' \mapsto (\gamma v) \otimes (\gamma v')$ is also semisimple.

Let $G \subset {\rm{GL}}(V \otimes V')$ be the Zariski closure of $r(\Gamma)$ (this is why I changed the original notation for the discrete group with which we begin). This is a smooth affine $k$-group, possibly disconnected, but we claim that we can say something non-trivial about the structure of $G$: it is reductive (in the sense that the unipotent radical of $G_{\overline{k}}$ is trivial).

Consider the natural map $$f:{\rm{GL}}(V) \times {\rm{GL}}(V') \rightarrow {\rm{GL}}(V \otimes V')$$ that satisfies $r = f \circ (\rho \times \rho')$. Let $H \subset {\rm{GL}}(V)$ and $H' \subset {\rm{GL}}(V')$ denote the respective Zariski closures of $\rho(\Gamma)$ and $\rho'(\Gamma)$, so $H \times H'$ contains the Zariski closure $\mathscr{H}$ of $(\rho \times \rho')(\Gamma)$. The magic begins with:

Lemma. The $k$-group $\mathscr{H}$ is a reductive group.

Proof: Under the projections $H \times H' \rightrightarrows H, H'$, the images of $\mathscr{H}$ are dense (contain $\rho(\Gamma), \rho'(\Gamma)$ respectively) and closed (as any homomorphism between smooth affine groups over a field has closed image), so both maps $\mathscr{H} \rightrightarrows H, H'$ are surjective. Recall a real theorem: under surjective homomorphisms between smooth affine groups over an algebraically closed field, the image of the unipotent radical of the course is the unipotent radical of the target. Thus, if we can show that $H$ and $H'$ are reductive then $\mathscr{R}_u(\mathscr{H}_{\overline{k}})$ would have trivial images in both $H_{\overline{k}}$ and ${H'}_{\overline{k}}$, yet $\mathscr{H}$ is a $k$-subgroup of $H \times H'$, so this would force $\mathscr{R}_u(\mathscr{H}_{\overline{k}}) = 1$ as desired.

So it suffices to show that $H$ and $H'$ are reductive, and this is exactly where we will use the assumptions that $\rho$ and $\rho'$ are semisimple. Let $U = \mathscr{R}_{u,k}(H)$ denote the so-called $k$-unipotent radical of $H$; i.e., the maximal smooth connected unipotent normal $k$-subgroup. Since $k$ is perfect (even char. 0), it follows by Galois descent that $U_{\overline{k}} = \mathscr{R}_u(H_{\overline{k}})$, so our task is equivalent to showing that $U=1$ (as the case of $H'$ goes in exactly the same way).

Let $V = \oplus V_i$ be the decomposition of $V$ into a direct sum of irreducible representations of $H$ over $k$. Since $U$ is unipotent, by the Lie-Kolchin Theorem the subspaces $V_i^U$ are each nonzero. But $U$ is normal in $H$, so each $V_i^U$ is an $H$-subrepresentation of $V_i$. Irreducibility of $V_i$ for $H$ then forces $V_i^U = V_i$, so $V^U=V$. In other words, $U$ acts trivially on $V$. But this action is defined through the inclusion $U \subset H \subset {\rm{GL}}(V)$, so $U=1$.

QED Lemma

Since $r = f \circ (\rho \times \rho')$, we have $G = f(\mathscr{H})$ because the image under any homomorphism between smooth affine groups (such as $f$) of a Zariski-closed subgroup of the source is a Zariski-closed subgroup of the target. But we also know that any quotient of a reductive group is always reductive (no connectedness hypotheses!), so the Lemma implies that $G$ is reductive (which is equivalent to that for its identity component $G^0$, since $\mathscr{R}_u(\mathscr{G}) = \mathscr{R}_u(\mathscr{G}^0)$ for general smooth affine groups $\mathscr{G}$ over any algebraically closed field and we recall that $(G^0)_{\overline{k}} = (G_{\overline{k}})^0$).

By design, the image of $\Gamma$ in $G(k)$ is Zariski-dense in $G$. Thus, a subspace of $V$ is $\Gamma$-stable if and only if it is $G$-stable. Thus, the semisimplicity of $V$ as a $\Gamma$-representation is equivalent to its semisimplicity as a $G$-representation. In view of the established reductivity of $G$ (equivalently, of $G^0$), now we are finally cooking with gas: we have transformed the original problem entirely into a question about the representation theory of (possibly disconnected!) reductive groups.

Up to here, we have only used that $k$ is perfect, not that it has characteristic 0. The special features of characteristic 0 now appear with the following fact that completes the proof:

Proposition Any finite-dimensional linear representation of a $($possibly disconnected$)$ reductive smooth affine group $G$ over a field $k$ of characteristic $0$ is completely reducible.

Proof: We need to show that of $W$ and $W'$ are finite-dimensional linear representations of $G$ over $k$ then any extension of $W$ by $W'$ as representations of $G$ is split. In symbols, we want the group ${\rm{Ext}}^1_{k[G]}(W,W')$ of isomorphism classes of such extensions to vanish.

Recalling that $G$-comodules over $k$ (possibly infinite-dimensional) are always a directed union of finite-dimensional algebraic subrepresentations of $G$, this category has enough injectives and so we can make sense of ${\rm{Ext}}^i$'s on this category. We'll show all higher Ext's vanish on finite-dimensional inputs. By a universal $\delta$-functor argument we have ${\rm{Ext}}^i_{k[G]}(W, \cdot) = {\rm{Ext}}^i_{k[G]}(k, W^{\ast} \otimes (\cdot))$. Since ${\rm{Ext}}^i_{k[G]}(k,\cdot)$ is the derived functor of formation of $G$-invariants, it suffices to show that the formation of $G$-invariants is right-exact.

Exhausting $G$-comodules by finite-dimensional subrepresentations, we are reduced to checking that if $W \twoheadrightarrow \overline{W}$ is a surjective map between finite-dimensional $G$-representations then $W^G \rightarrow \overline{W}^G$ is surjective. If $N \subset G$ is a smooth closed normal $k$-subgroup (so $N$ is also reductive, as is $G/N$) and we can settle the general right-exactness for $N$-invariants and $G/N$-invariants then we get the same for $G$. Moreover, since $(W^G)_{\overline{k}} = (W_{\overline{k}})^{G_{\overline{k}}}$, for the purpose of proving this right-exactness property we may and do assume $k$ is algebraically closed.

If $Z$ is the maximal central $k$-torus in $G$ then $G/Z$ has semisimple identity component $G^0/Z$ and $G/G^0$ is finite etale. Thus, we are reduced to separately treating three cases over algebraically closed fields of characteristic 0: $\mathbf{G}_m$, finite groups, and connected semisimple groups. The case of $\mathbf{G}_m$ works in any characteristic (as linear representations are the same as $\mathbf{Z}$-gradings), and the case of finite groups is well-known (in characteristic 0!).

Finally, we come to the real substance of the matter: connected semisimple $G$. The structure theory of such groups over $k$ of characteristic 0 tells us that the Lie algebra $\mathbf{g}$ is semisimple in the sense of Lie algebras, so the representation theory of $\mathfrak{g}$ is completely reducible in characteristic $0$.

It therefore suffices to show that if $W$ is a finite-dimensional linear representation of a connected linear algebraic group $G$ over $k$ of characteristic 0 (semisimplicity or even reducivity is not relevant anymore) and we also view $W$ as a linear representation of $\mathfrak{g}$ in the usual way (by passing to Lie algebras on the given homomorphism $G \rightarrow {\rm{GL}}(W)$) then $W^G = W^{\mathfrak{g}}$.

More generally, we claim that for any subspace $V \subset W$, (i) $V$ is $G$-stable if and only if $V$ is $\mathfrak{g}$-stable, and (ii) when those equivalent properties hold, the representation of $G$ on $V$ is trivial if and only if the $\mathfrak{g}$-representation on $V$ is trivial.

For the implication "$\Leftarrow$" in (i) (the converse being obvious), we want to show that if $H \subset {\rm{GL}}(W)$ is a Zariski-closed subgroup (such as the stabilizer of $V$) then a $k$-homomorphism $f:G \rightarrow {\rm{GL}}(W)$ factors through $H$ if and only if it does so on the level of Lie algebras. The scheme-theoretic preimage $f^{-1}(H)$ has Lie algebra that is the preimage of $\mathfrak{h}$ in $\mathfrak{g}$, so the closed $k$-subgroup scheme $f^{-1}(H) \subset G$ has full Lie algebra. But $f^{-1}(H)$ is smooth by Cartier's theorem (char. 0!), so by connectedness and smoothness of $G$ this forces $f^{-1}(H)=G$ by dimension considerations, which is to say that $f$ factors through $H$ as desired.

Assertion (ii) is a special case of the general assertion that any homomorphism between connected linear algebraic groups in characteristic 0 (such as $G \rightarrow {\rm{GL}}(V)$) is determined uniquely by the induced map between Lie algebras, and via consideration of graphs it reduces to showing that a Zariski-connected Zariski-closed subgroup in characteristic 0 is uniquely determined by the corresponding Lie subalgebra. This in turn in a fact proved in many books on algebraic groups (or one can appeal to the analytic theory over $\mathbf{C}$ provided one shows that Zariski-connectedness implies analytic connectedness, which requires a real argument and cannot be considered to be elementary).

QED