Let $G$ be a group, $S \subseteq G $ a subset that generates $G$, $R \subseteq F_S$, where $F_S$ is the free group on the set $S$.
Let $\psi: F_S \rightarrow G$ be a group homomorphism, s.t. $\psi(s) = s, \forall s \in S$ and $\psi(r) = e_G, \forall r \in R$
Suppose that $R$ is a maximal set of relators for $G$, meaning the following:
$\forall T \subseteq F_S, R \subseteq T,$ s.t. $\forall t \in T, \psi(t) = e_G,$ $\forall \phi:F_S \rightarrow H,$
with $H$ group and $\phi$ a group homormophism, then:
$$(\forall r \in R, \phi(r) = e_H) \Rightarrow (\forall t \in T, \phi(t) = e_H)$$
Then, under these conditions, is it true that the following holds?
$\forall \xi:F_S \rightarrow K$, with $K$ a group and $\xi$ s group homormophism, s.t. $\forall y \in F_S, \psi(y) = e_G \Longleftrightarrow \xi(y) = e_K $ then:
$\forall x \in F_S, \xi(x) = e_K \Longleftrightarrow x \in \langle R^{F_S} \rangle$
If true, is there a simple way to show it?
I think I understand what you are asking now.
If $r \not\in \langle R^{F_S} \rangle$, then $r$ maps onto a nontrivial element of the group $G:=F_S/\langle R^{F_S} \rangle$, and so $G$ is a group in which the words in $R$ all map onto the identity element, but $r$ does not.
So, if the assumption that all elements of $R$ map onto the identity of a group $H$ implies that $r$ maps onto the identity of $H$, then we must have $r \in \langle R^{F_S} \rangle$, which means that $r$ is equal in $F_S$ to a product of conjugates of elements of $R$.