We consider $\mathbb{Z}$-modules (i.e., abelian groups).
Since $\mathbb{Q}$ is divisible, if $A$ is a torsion abelian group, then $A\otimes\mathbb{Q}$ is trivial.
Let $G$ be the direct product of cyclic group of order $p^n$, with $p$ a prime, and $n$ increasing; that is:
$$G = \prod_{n=1}^{\infty}\mathbb{Z}/p^n\mathbb{Z}.$$
Then
$$\prod_{n=1}^{\infty}\left(\mathbb{Z}/p^n\mathbb{Z}\otimes\mathbb{Q}\right) = 0.$$
But $G\otimes\mathbb{Q}$ is not trivial: if we let $x$ be the element that corresponds to the class of $1$ in every coordinate, then $x$ has infinite order. Therefore,
$$\langle x\rangle \otimes\mathbb{Q}\cong \mathbb{Z}\otimes\mathbb{Q} \cong\mathbb{Q};$$
but tensoring with $\mathbb{Q}$ over $\mathbb{Z}$ is exact; therefore, the embedding $\langle x\rangle \hookrightarrow G$ induces an embedding $\langle x\rangle\otimes \mathbb{Q}\hookrightarrow G\otimes \mathbb{Q}$. Therefore, $G\otimes\mathbb{Q}\neq 0$. Thus, we have
$$\left(\prod_{n=1}^{\infty}\mathbb{Z}/p^n\mathbb{Z}\right)\otimes \mathbb{Q}\not\cong \prod_{n=1}^{\infty}(\mathbb{Z}/p^n\mathbb{Z}\otimes\mathbb{Q}).$$
Notation
$(G,\cdot_G)$ is a group with composition (or product) $\cdot_G$. The group of automorphisms of a vector space, let us say $V$, is denoted by $(\operatorname{Aut}(V),\circ)$, where $\circ$ is the composition of automorphisms.
The representations are defined in the OP; we use the following notation
$$\rho_1: (G,\cdot_G)\rightarrow (\operatorname{Aut}(V),\circ),$$
$$\rho_2:(G,\cdot_G)\rightarrow (\operatorname{Aut}(W),\circ).$$
We just recall that given any representation $\rho: (G,\cdot_G)\rightarrow (\operatorname{Aut}(T),\circ)$ we have
$$\rho(g_1\cdot_G g_2)=\rho(g_1)\circ\rho(g_2)$$
for all $g_1,g_2\in G$.
$$ \rho_\oplus:(G,\cdot_G)\rightarrow (\operatorname{Aut}(V\oplus W),\circ)$$
is given by $\rho_\oplus:=\rho_1\oplus\rho_2$, i.e.
$$\rho_\oplus(g)(v\oplus w)=\rho_1(g)(v)\oplus\rho_2(g)(w)\in V\oplus W$$ for all $v \in V$, $w\in W$ and $g\in G$.
By definition, it follows that $\rho_\oplus(g_1\cdot_G g_2)=\rho_\oplus(g_1)\circ\rho_\oplus(g_2)$ and $\rho_\oplus(g^{-1})=\rho^{-1}_\oplus(g)$. This makes $\rho_\oplus$ a group homomorphism. Let us prove the first one as example. The first equation is proven by
$$(\rho_\oplus(g_1\cdot_G g_2))(v\oplus w)=(\text{def. of}~\rho_\oplus)=
\rho_1(g_1\cdot_G g_2)(v)\oplus\rho_2(g_1\cdot_G g_2)(w)=(\text{def. of representations:})=
\rho_1(g_1)(\rho_1(g_2)(v))\oplus\rho_2(g_1)(\rho_2(g_2)(w))=
(\rho_\oplus(g_1)\circ\rho_\oplus(g_2))(v\oplus w);$$
the last equality follows from the definition of composition $\circ$ in $\operatorname{Aut}(V\oplus W)$. In fact:
$$(\rho_\oplus(g_1)\circ\rho_\oplus(g_2))(v\oplus w):=
\rho_\oplus(g_1)(\rho_\oplus(g_2)(v\oplus w))=(\text{def. of}~\rho_\oplus)=
\rho_\oplus(g_1)(\underbrace{\rho_1(g_2)(v)}_{\in V}\oplus \underbrace{\rho_2(g_2)(w)}_{\in W})=(\text{again def. of}~\rho_\oplus)=\\
\underbrace{\rho_1(g_1)(\rho_1(g_2)(v))}_{\in V}\oplus \underbrace{\rho_2(g_1)(\rho_2(g_2)(w))}_{\in W},$$
as wished.
$$ \rho_\otimes:(G,\cdot_G)\rightarrow (\operatorname{Aut}(V\otimes W),\circ)$$
is given by $\rho_\otimes:=\rho_1\otimes\rho_2$, i.e.
$$\rho_\otimes(g)(v\otimes w)=\rho_1(g)(v)\otimes\rho_2(g)(w)\in V\otimes W$$ for all $v \in V$, $w\in W$ and $g\in G$.
By definition, it follows that $\rho_\otimes(g_1\cdot_G g_2)=\rho_\otimes(g_1)\circ\rho_\otimes(g_2)$ and $\rho_\otimes(g^{-1})=\rho^{-1}_\otimes(g)$. This makes $\rho_\otimes$ a group homomorphism. The relations are proven in a similar way to the one used in the direct sum case.
Best Answer
For the best argument, see the comment by Qiaochu Yuan.
Here is an alternative argument: Let $R$ be a comutative ring. I assume that you already know that the tensor product of $R$-modules commutes with direct sums (see below for the isomorphism). Now let $G$ be a group and let $A,B,C$ be representations of $G$ over $R$, i.e. $R$-modules with an action of $G$. Then the isomorphism of $R$-modules
$f : (A \otimes B) \oplus (A \otimes C) \to A \otimes (B \oplus C), ~ f(a \otimes b)=a \otimes (b,0),~ f(a \otimes c)=a \otimes (0,c)$
is also $G$-equivariant, because $f(g(a \otimes b))=f(ga \otimes gb)=ga \otimes (gb,0)=ga \otimes g(b,0) = g f(a \otimes b)$ and likewise $f(g(a \otimes c))=g f(a \otimes c)$.
Actually, $f$ commutes with the $G$-actions simply because $f$ is a natural isomorphism. Even more abstractly, such a morphism $f$ exists in any monoidal category with coproducts, which is here the category ${}^G \mathsf{Mod}(R)$ of representations of $G$ over $R$. Therefore, actually no computation is needed at all.
Here is a generalization: Let $\mathcal{C}$ be a monoidal category and let $G$ be an arbitrary small category. Then the category ${}^G \mathcal{C}$ of functors $G \to \mathcal{C}$ is again monoidal, the tensor product is defined objectwise. If $\mathcal{C}$ has direct sums which commute with $\otimes$, then the same is true for ${}^G \mathcal{C}$, for trivial reasons: For $A,B,C \in {}^G \mathcal{C}$ there is a canonical morphism $(A \otimes B) \oplus (A \otimes C) \to A \otimes (B \oplus C)$ in ${}^G \mathcal{C}$. That it is an isomorphism, may be checked objectwise, and thereby reduced to $\mathcal{C}$.