As has been stated, $M rightarrow G$ factors via the group completion $hat{M}$. Furthermore, since $G$ is abelian, the map $hat{M} rightarrow G$ factors via the abelianization $hat{M}/[hat{M},hat{M}]$ of $hat{M}$. Since the composite of the maps $M rightarrow hat{M} rightarrow hat{M}/[hat{M},hat{M}]$ is already known, $f$ is uniquely determined by the map $hat{M}/[hat{M},hat{M}] rightarrow G$, so we may assume $M$ to be an abelian group.
Since $f = 0$ gives us no information, we may assume that $f$ is onto. Suppose we have a $G$ so that $f$ factors via it, and let us write $f = pg$, where $g colon M rightarrow G$ and $p colon G rightarrow {0,1}$.
By $f = pg$ we must necessarily have $ker(g) subset ker(f)$, so that $g$ is isomorphic to the canonical projection $M rightarrow M/N$, where $N$ is a subgroup of $ker(f)$. In addition, $|G|$ must be even. (If $f$ is onto, $p$ must take on the values $0$ and $1$ equally, since $|G| = 2|ker(p)|$.)
I claim this is all $f$ tells you. Given $f$, we've just shown that we must have that $g$ is isomorphic to modding out by a subgroup of $ker(f)$, and that $|G|$ is even.
Conversely, given any map $g colon M rightarrow G$, whose kernel is a subgroup of $ker(f)$, and such that $|G|$ is even, then the image of $ker(f)$ will be of index 2 in $G$, so we can just compose this map with the map that mods out the image of $ker(f)$ in $G$.
Edit: Oops, $G rightarrow {0,1}$ is only a set map! Well then, let me at least try to contribute to the discussion! In this case, we can still assume $M$ to be an abelian group. If you can recover $G$ (assuming $G = M$ is not allowed), then certainly the kernel of $M rightarrow G$ cannot have any subgroups, hence must be a cyclic group of order $p$. In the case where $M = mathbb{Z}_+$, then $hat{M} = mathbb{Q}_+$ which is torsion free, so no maps $f$ exist which allow you to recover $G$ entirely, unless we allow $G = hat{M}$.
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