The fact that $S$ is a monoid tells you that the operation is associative. You need to show that:
If $a,b$ have inverses, then $ab$ has an inverse. (Just produce the inverse using the inverses of $a$ and $b$). This shows you have a set and a binary associative operation on th eset.
Show that the operation has an identity; namely, the identity of $S$ lies in $S^{\times}$. (Show it has an inverse).
Show that if $a\in S^{\times}$, then there exists $b\in S^{\times}$ such that $ab=1$ (just produce it; you know $b$ exists in $S$ by definition of $S^{\times}$, prove it actually lies in $S^{\times}$.
This will show $S^{\times}$ is a group.
To prove that $U_n$ is a group using this, consider the monois $\mathbb{Z}_n$. Show that $a\in\mathbb{Z}_n$ has a multiplicative inverse if and only if $\gcd(a,n)=1$.
In one direction: if there exists $b$ such that $ab\equiv 1\pmod{n}$, then there exists $k$ such that $ab-1 = kn$, hence $ab-nk = 1$, so $\gcd(a,n)$ divides $ab-nk=1$. Thus, $\gcd(a,n)=1$. Try proving the other direction yourself.