Nobody stops you from making such a definition. However, what you get is not really useful. Let's see why.
Let $a\mathbin{\tau}b$ denote the element $(a,b)+N\in F/N$. Suppose $r,s\in R$; then, by definition,
$$
rs(a\mathbin{\tau}b)=r(sa\mathbin{\tau}b)=(rsa)\mathbin{\tau}b
$$
but also
$$
r(sa\mathbin{\tau}b)=(sa)\mathbin{\tau}(rb)=
s(a\mathbin{\tau}(rb))=
s(r(a\mathbin{\tau}b))=sr(a\mathbin{\tau}b)
$$
Thus
$$
rs(a\mathbin{\tau}b)=sr(a\mathbin{\tau}b)
$$
Hence $F/N$ is a module with the following property:
for each $x\in F/N$ and $r,s\in R$, $rsx=srx$.
When the ring is not commutative, this limits very much a possible usefulness of the concept.
For instance, suppose $A=B=R$. Then it is easily seen that $F/N$ is generated by $1\mathbin{\tau}1$, so there is a surjective module homomorphism $f\colon R\to F/N$ with $f(1)=1\mathbin{\tau}1$. The kernel is a left ideal $I$ such that $rs-sr\in I$, for every $r,s\in R$.
Now take the easiest noncommutative ring, the quaternions $\mathbb{H}$. The left ideal above is not $\{0\}$, so it is the whole ring and $F/N$ is the zero module. Similarly for the ring of $2\times2$ matrices over a field.
