Here is a counterexample. Let $k$ be a finite field, let $V=k^{\oplus\mathbb{N}}$, and let $W$ be a nonzero finite-dimensional vector space over $k$. Let $R$ be the ring of endomorphisms $f:V\oplus W\to V\oplus W$ such that $f(V)\subseteq V$. Let $a\in R$ be the endomorphism which is the left shift operator on $V$ and the identity on $W$. Let $b\in R$ be the endomorphism which is the right shift operator on $V$ and the identity on $W$. Let $d\in R$ be a nonzero endomorphism which is $0$ on $V$ and maps all of $W$ to the kernel of $a$ (i.e., the subspace of $V$ supported on the first coordinate).
Now observe that $ab=1$ and $ad=0$, so $r_2=b$ and $r_1=b+d$ are two different right inverses to $a$. For any $x\in R$, we then have $\gamma(x)=b+dx$, so the question is whether $dx$ takes infinitely many values. But any element of $R$ of the form $dx$ must annihilate $V$ and have image contained in the kernel of $a$, since $x$ maps $V$ to itself and $d$ annihilates $V$ and has image contained in the kernel of $a$. Since $W$ and the kernel of $a$ are both finite, there are only finitely many such maps. So $dx$ takes only finitely many different values, and so does $\gamma(x)$.
To relate your approach to the standard proof, note that if you fix one right inverse $r_2=b$, then you can get another right inverse $r_1=b+d$ by adding any $d\in R$ such that $ad=0$. One particular choice of such a $d$ is $d=1-ba$. The standard proof then shows that the elements $da^n$ are distinct for all $n\in\mathbb{N}$, so that $\gamma(x)=b+dx$ takes infinitely many different values. This only works, though, because of the special value of $d$ chosen (in particular, the argument uses the fact that $1-d\in Ra$). As the example above shows, if you pick an arbitrary nonzero $d$ such that $ad=0$, there may be only finitely many different values of $dx$.
