Let $x$ and $y$ be integers such that $xy = 0$. Then either $x = 0$ or $y = 0$ (or both).
MY ATTEMPT (EDIT)
Let us suppose that $x = (a,b)$ and $y = (c,d)$, where $a,b,c,d$ are natural numbers. We have that \begin{align*} xy = 0 \Longrightarrow (ac + bd, ad + bc) = (0,0) \Longrightarrow ac + bd = ad + bc \Longrightarrow (a = b)\vee(c = d) \end{align*}
If $a = b$, $x = (a,a) = (0,0)$, then $x = 0$. If $c = d$, then $y = (c,c) = (0,0)$, and $y = 0$. If $a = b$ and $c = d$, then $x = y = 0$, as desired.
Any comments or contributions to my solution?
EDIT
The set of integers has been defined as the set of equivalence classes of pairs of natural numbers according to the relation $(a,b)\sim(c,d)$ iff $a + d = b + c$. In this context, the sum has been defined as $(a,b) + (c,d) = (a+c,b+d)$ and the product has been defined as $(a,b)\times(c,d) = (ac + bd,ad + bc)$.
Your solution is not correct. In this context, $0$ is the equivalence class of $(0,0)$, which is $\{(n,n)\mid n\in\mathbb N\}$. So, what you should prove is that$$ac+bd=ad+bc\implies a=b\vee c=d.\tag1$$
EDIT
The new version of the proof is incomplete, since you did not prove $(1)$.
Suppose that $a>b$. Then $a=b+n$, for some $n\in\mathbb N\setminus\{0\}$. So, $(1)$ becomes$$\require{cancel}\cancel{bc}+nc+\bcancel{bd}=\bcancel{bd}+nd+\cancel{bc},$$which means that $nc=nd$. Since $n\neq0$, $c=d$. If $a<b$, a similar argument shows that, again, $c=d$.