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 $G_1 = G_2 = \mathbb{Z} / 2 \mathbb{Z}$.
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
Yes. Indeed, $V_1$ and $V_2$ are the only proper nontrivial subrepresentations. Any subreps would be sums of subirreps. If $W$ is any irrep, any homomorphism $W\to V_1\oplus V_2$ can be written as $\phi_1\oplus\phi_2$ for morphisms $\phi_1:W\to V_1$ and $\phi_2:W\to V_2$. By Schur's lemma, these two morphisms can only be trivial or isomorphisms, but they can't both be isomorphisms, so the only subirreps are $V_1$ and $V_2$, and any subrep must be these or a sum of these.
More generally, if $V$ is a direct sum of inequivalent irreps, then this decomposition is unique, and the only subreps are direct sums of these given subirreps.
This uniqueness fails when you start having more than one equivalent irrep as subreps. For instance, consider multiple copies of a trivial representation, $\mathbb{C}^n$. There are infinitely many trivial subreps of any intermediate dimension, and infinitely many decompositions into (1D) irreps.
However, there are unique "isotypic components" which contain all subreps isomorphic to a given irrep, they are the images of applying "isotypical projectors" from the group algebra (these elements kill any element of an irrep that is inequivalent to the given one, and fix any element in an irrep equivalent to the given one).