Let $G = \left\langle g_1, \ldots , g_n \mid R_1\right\rangle$ and $H = \left\langle h_1, \ldots , h_n \mid R_2 \right\rangle$.
Suppose there's a mapping $f$ s.t. $f(g_i) = h_i$. Can we then merely stipulate that $f$ can be extended to mapping $f'$ s.t. $f'(ab) = f'(a)f'(b)$ so that $f'$ is a homomorphism between $G$ and $H$?
If the relations are empty, this seems to be true. But what happens if the relations $R_1$ and $R_2$ aren't empty? Are there extra things to check or stipulate to determine whether $f'$ is a homomorphism?
You need more conditions. It works if you have for all $a_1\ldots a_r \in R,\; f(a_1)\ldots f(a_r) = 1_H$. (Where each $a_i = g_j^{\pm 1}$ for some $j$).
For example:
$$G = \langle x,y\; \mid\; x^3,\; y^3,\; xy\rangle \cong C_3$$ $$H = \langle u,v\; \mid\; u^2,\; v^2,\; uv=vu\rangle \cong C_2 \times C_2$$
If we define $f : G \rightarrow H$ by $f(x) = u,\; f(y) = v$, then you can check $f$ isn't a homomorphism.