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.
This does not work : why is $f$ equivariant ?
Given representations $\rho_1,\rho_2$ of $G$ on $V$, $id_V: (V,\rho_1)\to (V,\rho_2)$ is equivariant if and only if $\rho_1=\rho_2$ : this has nothing to do with Schur's lemma, or irreducibility, this is actually very easy to see.
Now you claim that $id$ is equivariant but to prove that you'd have to first prove that $\Gamma = \rho_1\otimes \rho_2$, which is what you want to prove anyway.
Note that your proof can't work anyway because you chose arbitrary $\rho_1,\rho_2$, so it's clear that it can't work.
Martin Brandenburg's answer in your related question proves this fact (he mentions it in his answer) in the case of an algebraically closed field of characteristic prime to $|G_1|,|G_2|$; if you can prove that $\rho_1\otimes \rho_2\cong \rho'_1\otimes \rho'_2 \implies \rho_1\cong \rho'_1 \land \rho_2\cong \rho'_2$.
However, Mariano Suarez-Alvarez already gives a proof (still in your related question) of your statement : he actually produces two irreducible representations $V_1,V_2$ of $G_1,G_2$ respectively whose tensor product is isomorphic to the $\Gamma$ you start with - you should definitely check out his answer, it's well-written !
The only point in it that is not entirely clear at a first glance is why $\hom_G(U,V_{\mid G}) $ has dimension $\leq \frac{\dim V}{\dim U}$, but that follows at once by using Schur's lemma and decomposing $V_{\mid G}$ into irreducible representations
(although to make sure that that holds you probably have to use some hypothesis, again something like the fact that the field is algebraically closed of characteristic prime to $|G|$)
Best Answer
When $V_1$ and $V_2$ are representations of $G_1$ and $G_2$ respectively, I'll use $V_1 \boxtimes V_2$ to mean the representation of $G_1 \times G_2$ with underlying vector space $V_1 \otimes_{\mathbb{C}} V_2$, and $V_1 \boxplus V_2$ to mean the representation of $G_1 \times G_2$ with underlying vector space $V_1 \oplus V_2$.
If $V$ is an irreducible representation of $G_1 \times G_2$, then $V$ is isomorphic to $V_1 \boxtimes V_2$ for some irreducible representations $V_1$ and $V_2$ of $G_1$ and $G_2$ respectively. This means that if we know the representations of $G_1$ and $G_2$, then using the $\boxtimes$ construction we can get to all the (irreducible) representations of $G_1 \times G_2$. Conversely, the $\boxtimes$ product of two irreducible representations always produces an irreducible representation of $G_1 \times G_2$.
On the other hand, $V_1 \boxplus V_2$ is always reducible as a $G_1 \times G_2$ representation, since both vector subspaces $V_1$ and $V_2$ are stable under the $G_1 \times G_2$ action. On the $V_1$ subspace, really only the $G_1$ part of the group acts, and the $G_2$ part acts trivially, and similarly for the $V_2$ subspace. We cannot produce all irreducible representations of $G_1 \times G_2$ using this construction, which can already be seen in the example $G_1 = G_2 = \mathbb{Z} / 2 \mathbb{Z}$.