$\newcommand{\eqclass}[1]{\left[#1\right]}$
Definition of equivalence relation on $\mathbb{N}^{2}= \mathbb{Z}$: for any $x, y$, $x \sim_{\mathbb{N}^{2}} y$ if and only if $x, y \in \mathbb{N}^2$ and for any $a, b, c, d \in \mathbb{N}$, if $x = \left(a, b\right)$, $y = \left(c, d\right)$, then $a + d = b + c$.
Equivalence class on $\sim_{\mathbb{N}^{2}}$: for any $x, a, b$, if $a, b \in \mathbb{N}$ and $x = \left(a, b\right)$, then for any $y$, $y \in \eqclass{x}$ if and only if $y \sim_{\mathbb{N}^2} x$.
Addition of two integers: for any $x, y \in \mathbb{Z}$ and $a, b, c, d \in \mathbb{N}$, if $x = \eqclass{\left(a, b\right)}$ and $y = \eqclass{\left(c, d\right)}$, then $x + y = \eqclass{\left(a + c, b + d\right)}$.
Multiplication of two integers: let $x, y \in \mathbb{Z}$ and $a, b, c, d \in \mathbb{N}$. If $x = \eqclass{\left(a,b\right)}$ and $y = \eqclass{\left(c, d\right)}$, then $x \cdot y = \eqclass{\left(ac + bd, ad + bc\right)}$.
Of course, $0_\mathbb{Z} = \eqclass{\left(a,a\right)}$ for any $a \in \mathbb{N}$.
I would like to prove the following lemma about integers as an equivalent class of ordered pairs of natural numbers: for any $x, y, z \in \mathbb{Z}$, if $x \neq y$ and $z \neq 0$, then $x \cdot z \neq y \cdot z$. My proof goes below:
Let $x, y, z \in \mathbb{Z}$ and assume that $x \neq y$ and $z \neq 0$. There are $a, b, c, d, e, f \in \mathbb{N}$ such that \begin{equation*} x = \eqclass{\left(a, b\right)}, \end{equation*} \begin{equation*} y = \eqclass{\left(c, d\right)}, \end{equation*} \begin{equation*} z = \eqclass{\left(e, f\right)}. \end{equation*} As $z \neq 0$. Then $e \neq f$. Further, as $x \neq y$, we have $a + d \neq b + c$. Further, \begin{equation*} x \cdot z = \eqclass{\left(a, b\right)} \cdot \eqclass{\left(e, f\right)} = \eqclass{\left(ae + bf, af + be\right)}. \end{equation*} \begin{equation*} y \cdot z = \eqclass{\left(c, d\right)} \cdot \eqclass{\left(e, f\right)} = \eqclass{\left(ce + df, cf + de\right)} \end{equation*} On the one hand, \begin{equation*} ae + bf + cf + de = \left(a + d\right) e + \left(b + c\right) f. \end{equation*} On the other hand, \begin{equation*} af + be + ce + df = \left(a + d\right)f + \left(b + c\right)e. \end{equation*} Without loss of generality, we assume $f < e$. Assume that $x \cdot z = y \cdot z$. Then \begin{equation*} ae + bf + cf + de = af + be + ce + df. \end{equation*} Immediately, \begin{equation*} \left(a + d\right) e + \left(b + c\right) f = \left(a + d\right)f + \left(b + c\right)e. \end{equation*} Then we have \begin{equation*} \left(a + d\right)\left(e - f\right) = \left(b + c\right)\left(e - f\right). \end{equation*} It is clear that $a + d = b + c$, leading to contradiction. Thus, $x \cdot z \neq y \cdot z$.
In my proof, I assumed $f < e$ and used subtraction of natural numbers. Nevertheless, it seems to me that in text books, we do not define $-$ on $\mathbb{N}$, as $-$ is not closed. How to prove the lemma without definition of subtraction?
To prove the $\sim_{\mathbb{N}^2}$ is an equivalence relation, we need to show that $\sim$ is reflexive, symmetric and transitive.
First of all, we show that $\sim$ is reflexive. Let $x \in \mathbb{N}^{2}$. There are $p, q \in \mathbb{N}$ such that $x = \left(p, q\right)$. Obviously, $p + q = p + q$. As a result, $x \sim x$.
Next, we show that $\sim$ is symmetric. Let $x \sim y$. Then there are $a, b, c, d \in \mathbb{N}$ such that $x = \left(a, b\right)$, $y = \left(c, d\right)$ and $a + d = b + c$. Obviously, $b + c = a + d$. Thus, $y \sim x$.
Finally, we show that $\sim$ is transitive. Let $x \sim y$ and $y \sim z$. By definition, there are $a, b, c, d, e, f \in \mathbb{N}$ such that $x = \left(a, b\right)$, $y = \left(c, d\right)$, $z = \left(e, f\right)$ and $a + d = b + c$ and $c + f = d + e$. Assume that $a + f \neq b + e$. Clearly $a + f + c + d \neq b + e + c + d$. On the other hand, we have $a + d + c + f = b + c + d + e$, leading to contradiction. Thus, $a + f = b + e$. (I used the following lemma without proof (has to be induction if proof is needed): for any $x, y, z\in \mathbb{N}$, x + y = x + z if only if $y = z$).
As $\sim$ is reflexive, symmetric and transitive on $\mathbb{N}^{2}$, $\sim$ is an equivalence relation on $\mathbb{N}^{2}$.
Aside: You shouldn't use contradiction to prove transitivity. You have a direct proof sitting inside it, so use it: If $(a,b)\sim (c,d)$ and $(c,d)\sim (e,f)$, then $a+d=b+c$ and $c+f=d+e$. From $a+d=b+c$, adding $f$ to both sides we get $a+d+f=b+c+f$. Since $c+f=d+e$, this gives $a+d+f=b+d+e$. Cancelling $d$, we get $a+f=b+e$, and this gives $(a,b)\sim(e,f)$.
You don't say whether your natural numbers include $0$ or not; mine do, but the argument below is easy to modify appropriately if yours do not.
The definition of order of natural numbers that I am aware of is:
In particular, $n\lt m$ if and only if $n\leq m$ and $n\neq m$, if and only if there exists a nonzero $h$ such that $n+h=m$.
So in order to avoid writing the difference, instead write that the larger number is a sum of natural numbers and use usual cancellation with natural numbers. So we are just expressing what we know in our hearts to be $e-f$ as $h$, and using that.
After $$\left(a + d\right) e + \left(b + c\right) f = \left(a + d\right)f + \left(b + c\right)e$$ we assume that $f\lt e$, and write $e=f+h$ for some nonzero natural number $h$. Substituting we get $$(a+d)e + (b+c)(e+h) = (a+d)(e+h) + (b+c)e$$ which gives $$(a+d)e + (b+c)e + (b+c)h = (a+d)e + (a+d)h + (b+c)e$$ and cancelling equal summands, we obtain $$(b+c)h = (a+d)h.$$ Since $h$ is nonzero, this would yield $b+c=a+d$, contradicting your hypothesis.
Alternatively, and better phrase, instead of writing it as a proof by contradiction, write it as a proof by contrapositive: if $z\neq 0$ and $xz=yz$, then $x=y$. The argument above shows ends with $b+c=a+d$, which gives $x=y$ as your conclusion.