Here's what I have.
Proof: For function $f$ to be a homomorphism, $f(x,y)f(a,b)=f((x,y) \ast (a,b))$, where $x,y,a,b \in \mathbb{Z}$ and $\ast$ is the operation of addition.
First let's look at $f((x,y) \ast (a,b))$.
$(x,y) \ast (a,b)=(x+a),(y+b)$ as an operation of addition of the group $\mathbb{Z} \times \mathbb{Z}$
$f((x+a),(y+b))=(x+a)-(y+b)$ as given by the definition of $f$.
Now let's look at $f(x,y)f(a,b)$.
$f(x,y)=x-y$ as given and $f(a,b)=a-b$ similarly.
$f(x,y)f(a,b)=(x-y)+(a-b)$ since it's given that the group operation is addition.
$f(x,y)f(a,b)=x-y+a-b=(x+a)-(y+b)$, since the group operation is addition in $\mathbb{Z}$. We know that commutative and associative properties hold.
Therefore, we have $f(x,y)f(a,b)=f(x,y \ast a,b)$ and the function is a homomorphism.
Q.E.D.
What I'm unsure about is if I can manipulate the final equation to match the first in the way that I did.
The proof looks correct. I would prove it more compactly instead of doing both sides separately:
Let $(a,b)$ and $(c,d)$ be in $\mathbb{Z}^2$. Then
$$ \begin{aligned} f((a,b)+(c,d)) &= f((a+c,b+d))\quad\text{definition of + on $\mathbb{Z}^2$}\\ &=(a+c)-(b+d)\quad\text{definition of $f$}\\ &=(a-b)+(c-d)\quad\text{$(\mathbb{Z}, +)$ is abelian and associative and inverse axioms of group $\mathbb{Z}$} \\ &= f((a,b))+f((c,d))\quad\text{definition of $f$} \end{aligned} $$
Your notation at present is kind of confusing. For example,
$(x,y)∗(a,b)=(x+a),(y+b)$ is a bit disorienting to me because it looks almost like you're mapping two ordered pairs into two objects. I know it's actually one ordered pair but the left side encloses the ordered pairs with parentheses, while the right side doesn't. So I would personally convert the asterisks into $+$'s to make it easier to read and make sure to enclose every ordered pair with parentheses to make it be slightly less ambiguous.