If $ab$ is an element of $G$ and $a\ast b=b\ast a$, we say that $a$ and $b$ commute. Suppose 3 and 2 commute. Does $3^{-1}$ and $2^{-1}$ commute I think it will also commute because it is under multiplication.
Yes they are, since (I took $a=3$ and $b=2$ for clearer presentation)

$a^{-1}b^{-1} = (ba)^{-1} = (ab)^{-1} = b^{-1}a^{-1}$

The point here is the formula $(ab)^{-1}=b^{-1}a^{-1}$ (you can check it by hand just applying the definition of inverse and using its unicity) that allows you to apply the hypothesis of commutation on $a$ and $b$.
If $a*b = b*a$ (i.e. $a$ and $b$ commute), then $a*b$ and $b*a$ have the same inverse.  In other words, $(a*b)^{-1} = (b*a)^{-1}$.  But then, by properties of the inverse, this is

$b^{-1}*a^{-1} = (a*b)^{-1} = (b*a)^{-1} = a^{-1}*b^{-1}$

By a similar argument, it can be shown that

$a$ and $b$ commute if and only if $a^{-1}$ and $b^{-1}$ commute.

That being said, without knowing what $G$ is, it is unclear if this is the statement you need or not.