From the apparent lack of complete answers in the literature, I figured I'd write the whole proof of the well-definedness of multiplication since it is only boring, not hard. I will write the equivalence class of a fraction $as^{-1}$ as $(a,s)$; and the equivalence relation as $\sim$. I'm assuming that you have full proofs that $\sim$ is a well-defined equivalence relation, and two classes can be brought to a common denominator. (For instance, you can find such proofs in Lam's "Lectures on Modules and Rings".)

Given classes $(a_1,s_1)$ and $(a_2,s_2)$, by the Ore condition we have $s_1 r=a_2s$ for some $r\in R$ and $s\in S$ (where $S$ is the multiplicative set we are inverting). Then we want to define $(a_1,s_1)*(a_2,s_2)=(a_1r,s_2s)$.

**Claim 1:** This is independent of the choice of $r$ and $s$.

Indeed, suppose we also have $s_1 r'=a_2s'$ for some $r'\in R$ and $s'\in S$. Since $(a_1r,s_2s)=(a_1rx,s_2sx)$ for any $x\in R$ such that $sx\in S$ (in fact, more generally for any $x\in R$ such that $s_2sx\in S$) from how we defined the equivalence classes, after replacing $s$ and $s'$ by a common denominator, we may as well assume $s'=s$. Then $s_1r=a_2s=a_2s'=s_1r'$, hence $s_1(r-r')=0$. By the zero-divisor condition, $(r-r')t=0$ for some $t\in S$. Thus $(a_1r,s_2s)\sim (a_1rt,s_2st)\sim (a_1r't,s_2s't)\sim (a_1r',s_2s')$.

**Claim 2:** Multiplication is independent of the choice of representative for the class $(a_1,s_1)$.

It suffices to prove this for another representative of this class of the form $(a_1x,s_1x)$ for some $x\in R$ with $s_1x\in S$. (Why? Because any two representatives can be brought to a common denominator.) The Ore condition gives us $(s_1x) r'=a_2s'$ for some $r'\in R$ and $s'\in S$.

By Claim 1, we may as well assume $s=s'$. Then $s_1xr=a_2s'=a_2s=s_1r$, hence $s_1(r-xr)=0$. By the argument at the end of claim 1 we are done with claim 2.

**Claim 3:** Multiplication is independent of the choice of representative for the class $(a_2,s_2)$.

Again, it suffices to do this for a representative $(a_2x,s_2x)$ for some $x\in R$ with $s_2x\in S$. By the Ore condition, write $sr'=xs'$ with $r'\in R$ and $s'\in S$. (This is a little different than the other uses of the Ore condition.) Then $s_1rr'=a_2sr'=a_2xs'$.

We just need to see that $(a_1r,s_2s)\sim (a_1rr',s_2xs')$. This is obvious, since if we multiply the first set of reps on the right by $r'$ we get the second set of reps.