If the hash of the multiplication is equal to the multiplication of the hash, how can this be used to leverage an attack? Assume a hash function $H:\left\{0,1\right\}^*\to G$ where $G$ is a group and assume that finding an inverse in $G$ is easy.

How can a preimage efficiently be found using the fact that $H(M_1\cdot M_2)=H(M_1)\cdot H(M_2)$ for $M_1,M_2\neq0$?

I've tried playing around with the group properties but I haven't gotten very far. Any direction would be appreciated.
Perform a birthday attack:

Given $y\in G$ we want to find $w\in \left\{ 0,1 \right\}^*$ such that $H\left(w\right)=y$.

Randomly choose $x\in \left\{ 0,1 \right\}^*$ and calculate $H\left(x\right)$ and $yH\left(x\right)^{-1}$. Wait until we find $a,b$ such that $H(a)$=$yH\left(b\right)^{-1}$.

Now $w=ab$: $H(w)=H(ab)=H(a)H(b)=yH\left(b\right)^{-1}H(b)=y$.