I've shown that $\phi^{-1}(\langle S'\rangle ) \supseteq \langle \phi^{-1}(S')\rangle$ and was wondering whether the other inclusion holds, although I've not been able to prove it nor find a counterexample.
I case it helps, one may use the fact that if $S\subseteq G$, then $$\phi(\langle S\rangle) = \langle \phi(S)\rangle.$$
On
For your question: The answer is FALSE.
It seems that you need some extra condition on $\phi$ (e.g., $\phi$ is surjective). If $\phi$ is not surjective, we can choose a non-empty set $S'\subseteq G'$ such that $\phi^{-1}(S')$ is empty. Then $\langle\phi^{-1}(S')\rangle$ is the trivial group $\{e_{G}\}$.
On the other hand, $\langle S'\rangle$ may be big, then $\phi^{-1}(\langle S'\rangle)$ is non-trivial.
Counter-example: Let $G=\{0,3,6,9\}$ and $G'=\mathbb{Z}_{12}$. Here, the group law in $G$ is: $(m,n)\mapsto m+n\,\,(\mbox{mod}$12). Let $\phi:G\rightarrow G'$ be defined by $\phi(x)=x$. Then $\phi$ is a homomorphism. Let $S'=\{2\}$. Then $\phi^{-1}(S')=\emptyset$, so $\langle\phi^{-1}(S')\rangle=\{0\}$.
On the other hand, $\langle S'\rangle=\{0,2,4,6,8,10\}$. Therefore, $\phi^{-1}(\langle S'\rangle)=\{0,6\}$
Hints: Let $S'_{0}=S'\cup\{a^{-1}\mid a\in S'\}$. We go to prove that $\langle S'\rangle=\cup_{n\in\mathbb{N}}\{a_{1}a_{2}\ldots a_{n}\mid a_{i}\in S'_{0}\}$. Clearly $RHS\subseteq\langle S'\rangle$ and $RHS$ contains $S'$. We verify that $RHS$ is a subgroup of $G'$. Let $x,y\in RHS$. Write $x=a_{1}\ldots a_{m}$ and $y=b_{1}\ldots b_{n}$ for some $a_{i},b_{j}\in S'_{0}$. We have that $xy=a_{1}\ldots a_{m}b_{1}\ldots b_{n}\in RHS$ and $x^{-1}=(a_{m})^{-1}\ldots(a_{1})^{-1}\in RHS$. Since $\langle S'\rangle$ is the smallest subgroup of $G'$ containing $S'$, it follows that $RHS\supseteq\langle S'\rangle$.