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 $|G|$)
Surely this $\tau:V_1\to V_2$ in your definition of "similar" should be a (linear) isomorphism as well. Otherwise any two representations with the same dimension would be similar (note that the trivial linear map $\tau(x)=0$ always satisfies the similarity condition). And thus your claim would be trivially true.
Anyway, by the assumption we have a representation isomorphisms $\tau:V_1\to V_2$ and $\theta:V_3\to V_4$. Those linear isomorphisms induce a linear map:
$$\gamma:V_1\oplus V_3\to V_2\oplus V_4$$
$$\gamma(x,y)=\big(\tau(x), \theta(y)\big)$$
It is easy to see that it is a linear isomorphism (the inverse is of the same form).
It also satisfies the conditions for similarity:
$$(\rho_1\oplus\rho_3)(\gamma(x,y))=(\rho_1\oplus\rho_3)(\tau(x),\theta(y))=$$
$$=\rho_1(\tau(x))\oplus \rho_3(\theta(y))=\tau(\rho_2(x))\oplus\theta(\rho_4(y))=$$
$$=(\tau\oplus\theta)\big((\rho_2\oplus\rho_4)(x,y)\big)=\gamma\big((\rho_2\oplus\rho_4)(x,y)\big)$$
and thus
$$(\rho_1\oplus\rho_3)\circ\gamma=\gamma\circ(\rho_2\oplus\rho_4)$$
which completes the proof.
Note that $V_i$ being of finite dimension is irrelevant.
Best Answer
In the case that only $V_2$ is irreducible, you can prove surjectivity by applying the argument proving Schur's lemma to the cokernel of $\varphi$. Similarly when $V_1$ is irreducible, you can apply the argument to the kernel of $\varphi$.
Edit: Actually, these are already the two pieces of the proof of Schur's lemma if you examine exactly where you use the irreducibility hypothesis on $V_1$ and $V_2$.