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.)
Best Answer
Your proof is okay, but it can be improved.
Closed: For any $l,m \in \mathbb{E}$ (where $\mathbb{E}$ is the set of even integers) $l = 2a$ and $m = 2b$ for some $a,b \in \mathbb{Z}$. Thus, $l + m = 2a + 2b = 2(a+b) \in \mathbb{E}$.
Identity: $0 \in \mathbb{E}$ and $0 + m = m + 0 = m$ for all $m \in \mathbb{E}$.
Inverse: For any $2k \in \mathbb{E}$, the number $2(-k) \in \mathbb{E}$ and $2k + 2(-k) = 2k - 2k = -2k + 2k = 0$.
Associative: The integers under addition are associative, so $\mathbb{E} \subset \mathbb{Z}$ inherits that property.