Let $$G= \langle a_i, i\in I\mid R_j, j \in J\rangle$$
Suppose we have a group $H$ with a subset $\{b_i\}_{i \in J}$ such that these elements satisfy the same relations as the elements $\{a_i\}_{i \in I}$.
I'm trying to prove that there is a unique homomorphism
$$\psi: G \to H$$
such that $\psi(a_i) = b_i$ for all $i \in I$. Is the following correct?
Attempt: Uniqueness:
Suppose $\psi: G \to H$ is a morphism such that $\psi(a_i) = b_i$ for all $i \in I$. Then the image of every element $x$ is already determined, because we can write every element as a product of $\{a_i\}_{i \in I}$ and their inverses and we can apply the homomorphism property to calculate the image in $H$.
Existence:
Define
$$\psi(a_{i_1}^{\epsilon_1}\dots a_{i_n}^{\epsilon_n}) = b_{i_1}^{\epsilon_1}\dots b_{i_n}^{\epsilon_n}$$
where $\epsilon_i \in \{-1,1\}$.
We first show that this map is well defined.
Suppose $$a_{i_1}^{\epsilon_1}\dots a_{i_n}^{\epsilon_n}= a_{k_1}^{\epsilon'_1}\dots a_{k_m}^{\epsilon'_n}$$
Then
$$a_{i_1}^{\epsilon_1}\dots a_{i_n}^{\epsilon_n}a_{k_m}^{-\epsilon'_n}\dots a_{k_1}^{-\epsilon'_1}=1 $$
Thus in $G$ we have the above relation, which can be derived from the given relations. Because the elements $b_i, i \in I$ satisfy the same relations in $H$, we can also derive the following relation in $H$
$$b_{i_1}^{\epsilon_1}\dots b_{i_n}^{\epsilon_n}b_{k_m}^{-\epsilon'_n}\dots b_{k_1}^{-\epsilon'_1}=1 $$
and hence the function is well defined.
Clearly the map is a homomorphism and we are done.
Is this correct?
Your proof is correct (though perhaps you should say an additional word why $\psi$ is clearly a homomorphism).
Here is an alternative proof. The group has a presention given by generators $a_i$ and relations $R_j = 1$. Each relation is obtained from a formal expression $\overline{R}_j(x_1,\epsilon_1,\dots,x_n,\epsilon_n) = x_1^{\epsilon_1}\dots x_n^{\epsilon_n}$ by inserting suitable $a_i$, i.e. we have $R_j = \overline{R}_j(a_{i(j,1)},\epsilon_{(j,1)},\dots,a_{i(j,n(j))},\epsilon_{(j,n(j)})$. In other words, we have $G = F/N$ where $F$ is the free group $F$ with generators $a_i$ and $N$ is the normal subgroup generated by the $R_j$. Now there exists a unique homomorphism $\phi : F \to H$ such that $\phi(a_i) = b_i$. Since the $b_i$ satisfy the same relations as the $a_i$ (i.e. $\overline{R}_j(b_{i(j,1)},\epsilon_{(j,1)},\dots,b_{i(j,n(j))},\epsilon_{(j,n(j)}) = 1$ in $H$), we see that $\phi(n) = 1$ for all $n \in N$. Hence $\phi$ induces a unique homomorphism $\psi : G = F/N \to H$ such that $\psi \circ p = \phi$ with $p : F \to F/N$ the quotient homomorphism (see Universal property of the quotient group.). But now any homomorphism $\chi : G \to H$ such that $\chi(a_i) = b_i$ satisfies $\chi \circ p = \phi$ and we conclude $\chi =\psi$.