To simplify notation, I will state the question in a less general form. I suspect the answer I'm looking for will be easily extended to the general case that I'm interested in.
Let $S$ be a collection of symbols and $u,v$ be two words in $S \sqcup S^{-1}$. Define two groups via the presentations $$G = \langle S \mid u,v \rangle \text{ and }H = \langle S \mid uv \rangle.$$ I want to show that $G$ is not isomorphic to $H$. I've tried using the fact that there's an epimorphism $H \to G$ which extends the inclusion map $S \hookrightarrow G$, but I haven't been able to proceed from this.
Here is a less trivial counterexample. A non-Hopfian group is one that is isomorphic to a proper quotient of itself.
Well known examples include the Baumslag-Solitar groups $${\rm BS}(m,n) = \langle x,y \mid y^{-1}x^myx^{-n} \rangle$$ for various values on $m$ and $n$.
In particular, ${\rm BS}(2,3)$ is non-Hopfian and if we let $w=y^{-1}x^2yx^{-3}$ and $v = (x^{-1}y^{-1}xy)^2x^{-1}$, then $$\langle x,y \mid w \rangle \cong \langle x,y \mid w,v \rangle.$$ So we get a counterexample to your question by taking $u=wv^{-1}$.
I was wondering whether the natural epimorphism $\langle S \mid u,v \rangle \to \langle S \mid uv \rangle$ has nontrivial kernel except in a few obvious cases, but that will not work either.
If the relator $v$ is a consequence of the relator $w$, then we can do the same thing. For example, we could take $w = [x,y]$ and $v = [x^2,y]$.