Uniquness of multiplicative inverses in $\Bbb Z_n$ (or any abeliean monoid) 
Assume that an integer $a$ has a multiplicative inverse modulo an integer $n$. Prove that this inverse is unique modulo $n$.

I was given a hint that proving this Lemma:
\begin{align}
n \mid ab \  \wedge \  \operatorname{gcd}\left(n,a\right) = 1
\qquad \Longrightarrow \qquad
n \mid b
\end{align}
should help me in finding the answer.
Here are my steps in trying to solve the problem:
\begin{align}
\operatorname{gcd}\left(n,a\right) = 1
& \qquad \Longrightarrow \qquad
sn + ta = 1
\qquad \Longrightarrow \qquad
sn = 1 - ta \\
& \qquad \Longrightarrow \qquad
1 \equiv ta \mod n
\qquad \Longrightarrow \qquad
ta \equiv 1 \mod n .
\end{align}
I know that having the GCD of m and a equal to 1 proves there is a multiplicative inverse mod n, but I'm not sure on how to prove $n \mid b$ with and how it helps prove the multiplicative inverse is unique.

What you have shown already is a technique that, using the Extended Euclidean Algorithm will give you the inverse (if it exists). Note also that you can determine if an element $a \mod n$ has an inverse by checking that they have no common divisors (that is, $\gcd(a,n)=1$).
Now to answer your question. Suppose that you have two multiplicative inverses $b,c$ such that $ab\equiv 1 \equiv ac \mod n$. (Then you also have $ba\equiv 1 \equiv ca \mod n$) Consider
$b \equiv b\cdot 1 \equiv bac \equiv 1\cdot c \equiv c \mod n$.
That shows that modulo $n$, there exists a unique inverse. This technique is very useful and you can also use it to show that: there exists exactly one $0$-vector in every vector space, group, ring... and every invertible element in a group, ring... has a unique inverse.
Note: For instance $2\cdot 3 \equiv 1 \mod 5$, but also $2\cdot 8 \equiv 1 \mod 5$ (Why is this not a contradiction to the above?)
