I know the answer to the question above is "no", quite flatly. The counter example is below:
$$\mathbb{Z}\cong\langle a,b\mid b^2a^{-1}\rangle\cong \langle a,b\mid\lbrace b^{2^{n+1 }}a^{-2^n}:n\in\mathbb{N}\rbrace\rangle.$$
So my question, more broadly, is that if $\langle A|R\rangle\cong\langle A|R'\rangle$ and we know that $R$ is finite, what can we say about $R'$?
I think we can say that $\langle R\rangle^N\cong \langle R'\rangle^N$, viewed inside the free group on $A$, but I'm not sure.