You've answered your own question. You say you're familiar with the fact that $(-b)c = -bc$ and $b(-d) = -bd$ in a ring. Now, if you write down a definition of multiplication that satisfies all of those properties, you get a ring structure on the set of equivalence classes, and $(-b)c = -bc$ and $b(-d) = -bd$ in a ring...!
Perhaps some extra generality will make this argument seem less circular. You have a rig $R$. This rig has an underlying additive monoid $M = (R, +)$. This abelian monoid has a Grothendieck group $G(M)$, which is the universal group into which $M$ maps. More precisely, there is a forgetful functor $F : \text{Ab} \to \text{CMon}$ from the category of abelian groups to the category of commutative monoids, and the Grothendieck group functor is its left adjoint. $G(M)$ consists of equivalence classes of formal expressions $m - n$ where $m, n \in M$ with addition defined as expected.
Now you want to put a ring structure on $G(M)$ compatible with the multiplication on $R$. It turns out that what you are actually constructing is the universal ring into which $R$ maps; more precisely, there is a forgetful functor from the category of rings to the category of rigs and you are constructing its left adjoint. Well, in any such ring we need
$$(a - b)(c - d) = ac - bc - ad + bd$$
as you well know, so this is the unique possible ring structure on $G(M)$ compatible with the multiplication on $R$ (and distributivity, inverses, etc.) if it is well-defined. This is a general property of adjoint functors (they are unique up to unique isomorphism if they exist). The only thing left is to verify that it actually works.
Another way to say the above is the following. Let $R$ be a rig and let $S$ be a ring, and let $\phi : R \to S$ be any rig homomorphism whatsoever. Then
$$(\phi(a) - \phi(b))(\phi(c) - \phi(d)) = \phi(ac) - \phi(bc) - \phi(ad) + \phi(bd).$$
There is nothing circular about this because $S$ is a ring and the above computation takes place in $S$.
Using $(m,n)\mapsto\dfrac{(m+n)(m+n+1)}2+m$, we can go the other way round by considering $k=T(n)+m$, where $T(n)=\dfrac{n(n+1)}{2}$, $n$ is maximal and $m\ge0$ to give $(m,n-m)$ as the pairing.
Best Answer
Yes, this is possible. Consider the relation $(a,b) \sim (c,d) \text{ iff } a+d = c+b$.
Verify that $\sim$ is an equivalence relation and let $[(a,b)]$ be the equivalence class of $(a,b)$ with respect to $\sim$. Define
$[(a,b)] +_\sim [(c,d)] := [(a+c,b+d)]$ and $-_\sim[(a,b)] := [(b,a)]$.
Prove that these are well-defined functions and that $$\pi \colon (\{ [(a,b)] \colon a,b \in \mathbb N \}, +_\sim) \to (\mathbb Z , +), [(a,b)] \mapsto a-b$$ is an group isomorphism.
I leave it to you to define $\cdot_\sim$ such that $\pi$ becomes a ring isomorphism.