Prove that $(\mathbb{Z}_n , +)$, the integers $\pmod{n}$ under addition, is a group.

A first important point: the elements of $\Bbb Z_n$ are not integers, but equivalence classes of the modulo $n$ relation. Some texts write these elements as $[0],[1],[2],\dots, [n-1]$ to emphasize they're equivalence classes (and hence sets of integers), where $[k]=\{x \in \Bbb Z: x\equiv k \pmod{n}\}$. In other words, the equivalence class of $k$ under this relation is the set of integers that leave a remainder $k$ when divided by $n$.

So you have $[a]=[b] \iff a\equiv b \pmod{n}\iff n \mid (a-b)$.

On these equivalence classes, you define addition as follows:

$$[a]+[b]=[a+b].$$

A way to become familiar with the operation is to check that it is well-defined, or to check that if $[a]=[c]$ and $[b]=[d]$, then $[a+b]=[c+d]$. (Try to check this, because the proofs of the group axioms are pretty much entirely based on the operation.)

As an example let's look at $\Bbb Z_3=\{[0],[1],[2]\}$.

Here $$[0]=\{\dots,-6,-3,0,3,6,\dots\} \\ [1]=\{\dots,-5,-2,1,4,7,\dots\} \\ [2]=\{\dots,-4,-1,2,5,8,\dots\}$$

We can write out the multiplication table (dropping the brackets for legibility):

\begin{array}{c|ccc} + & 0 & 1 & 2 \\ \hline 0 & 0 & 1 & 2 \\ 1 & 1 & 2 & 0 \\ 2 & 2 & 0 & 1 \end{array}

The table tells you that $[0]$ is the identity, and $[1]$ and $[2]$ are inverses. Note that $[1]+[2]=[1+2]=[3]=[0]$, where $[3]=[0]$ because here $n=3$.

Now, let's look at your proof.

a) Associative Law: First, assume that n is a composite number and that $a,b\in \Bbb Z$ such that $a=nx$ and $b=ny$ for some integers, $x$ and $y$. Then $a+nx+nx=n(x+y)$.

Is there a reason why you're assuming $n$ is composite? $\Bbb Z_n$ is a group also when $n$ is prime.

In your proof, by assuming that $a$ and $b$ are multiples of $n$ you're actually assuming they are elements of the equivalence class $[0]$, because $[nx]=x[n]=x[0]=[x0]=[0]$.

What you're required to show for the associative law, is that for three elements $[a],[b],[c]$ of $\Bbb Z_n$ (not integers, but equivalence classes) you have $[a]+([b]+[c])=([a]+[b])+[c]$.

The proof should follow from the way the operation of $+$ is defined for these equivalence classes and from associativity of addition of integers.

b) Existence of Identity: Generally, when showing the existence of an identity we can use this information for addition : a+0=0+a=a. I am unsure how to insert the modular portion of the definition into the existence of this identity. I realize that for addition, we generally look at 0 (rather than 1 for multiplication).

You want to show that $[0]=\{x \in \Bbb Z : x \equiv 0 \pmod{n}\}$, the equivalence class of all multiples of $n$, is the identity of $\Bbb Z_n$, or that for any $[a]\in \Bbb Z_n$ you have

$$[a]+[0]=[0]+[a]=[a]$$

This also follows from the definition of the operation: $[a]+[b]=[a+b]$.

c) Existence of Inverse: I need to show that for each element $a\in \Bbb Z_n$ that $a^{−1}=−a$. Again, I do not know how to insert the modular portion of definition into this aspect of the proof.

Since the operation here is addition, not multiplication, the inverses should be the negatives: try to show that the inverse of $[a]$ is $[-a]$ and that $[-a]$ is an element of $\Bbb Z_n$.

(I can give more details if you need them.)