Let $(G,*)$ be a group.
Let $S$ be a subset of $G$.
Then, construct the free group $(F(S),*')$ on $S$.
If there exists an isomorphism $\phi:(G,*)\rightarrow (F(S),*')$ such that $\phi(s)=(s)$ on $S$, then we say "$(G,*)$ is free on $S$".
Is my understanding correct? I'm asking this because several books have distinct definitions(i.e. fraleigh and Dummit&Foote)
I would read that as correct. As another way of looking at it: If $G$ is a free group generated by the set $S$, then the universal map from $F(S)$ to $G$ will be a surjective isomorphism. The inverse of this map will exactly be the map $\phi$.
Granted, it is possible for $G$ to be free without this being satisfied, depending on how you choose $S$: For example, let $G = \mathbb{Z}$, and let $S = \{2\}$. Then the free group on $S$ is just $\mathbb{Z}$ again, and the universal map $F(S) \to G$ is the inclusion $2\mathbb{Z} \hookrightarrow \mathbb{Z}$. This of course does not have an inverse.
This doesn't contradict anything that you've said, of course; $\mathbb{Z}$ is not free on the set $\{2\}$. More an idle comment, I guess.