Equality of fractions in localization as quotient Let $S\subset A$ be a subset of a commutative ring. Consider the polynomial algebra $A[x_s\mid s\in S]$ and define the localization $A_S$ as the quotient by the ideal $ (sx_s-1\mid s\in S)\vartriangleleft A[x_s\mid s\in S]$.
Let us write $\frac as$ for the equivalence class of $ax_s$ in the quotient $A_S$. I would like to prove that $\frac as=\frac bt$ iff $\exists u\in S$ such that $u(at-bs)=0$, like in the construction of localization as a quotient of $A\times S$.
So we have $\frac as-\frac bt=0\in A_S$ which is equivalent to $ax_s-bx_t\in I$. By degree considerations I think we must have $ax_s-bx_t=c(sx_s-1)-c(tx_t-1)$ for some $c\in A$, and by comparing coefficients we have $a=cs,b=ct$. Multiplying by $t,s$ respectively we find $at=cst=cts=bs$, whence $at-bs=0$ without any $u\in S$.
What is my mistake?
 A: I think the main mistake is in the statement "by degree considerations..." I did not verify that it is true for integral domains, but these considerations are more complicated in the presence of zerodivisors.
Let's consider a (classical) localization in a ring $A$ with zerodivisors. For example, let $A = \mathbb{Z}[y]/(2y)$ and invert $2$. Then in the localization $y=0$.
Now consider your polynomial-ring construction, in which $2x_2 - 1 = 0$ and (multiplying by $y$) $$2x_2 y - y = 0,$$ i.e. $y=0$. This looks like a separate, more minor, problem: it is a relation not of the form $$\frac{a}{s}=\frac{b}{t}$$ for $s,t \in S$ (you might want to take $1 \in S$.) I'll assume we took $1\in S$, and use a corresponding variable $x_1$ below.
Anyway, $y=0$ and in particular we have for example $$yx_2 - 0x_1 = yx_2 \in I.$$ So by your degree considerations, we should get $yx_2 = c(2x_2 - 1) - c(1x_1-1)$ in $A[x_1, x_2]$ for some $c\in A$. This is not the case. To exhibit this relation as an element of $I$ you need something like $yx_2 \cdot (2x_2 - 1),$ where $yx_2$ is not of degree $0$, hence not in $A$, so the degree considerations fail.

I previously wrote a "correction" to your proof, but it was mainly nonsense. There was some germ of a good idea hidden there, though. Here is a sketch of a correct proof:
Consider a ring $A$ and a module $M$ over it, and take $$(M\oplus M) / ((x,sx))$$ for some fixed $s\in A$. What is the submodule generated by the first coordinate? If an element $x$ is annihilated by $s$, then $$(x,0) = (x,0) - (x,sx) = (0,0)$$ in the quotient. It can be verified directly that the submodule is just $$\frac{M}{\text{ker}(s:M\rightarrow M)}.$$ What if we take $$\bigoplus_{i,j \ge 0} M$$ and mod out by elements of the form $$se_{i+1,j}-e_{i,j},\quad te_{i,j+1}-e_{i,j}?$$ I claim the $e_{0,0}$ coordinate then generates a module isomorphic to $$\frac{M}{\bigcup_n \text{ker}(st)^n:M\rightarrow M},$$ et cetera.
To see this, consider the colimit explicitly. If an element of the submodule generated by $M_{0,0}$ in the quotient vanishes, it vanishes already in some f.g. submodule, which is contained in some direct sum of a finite set of $M_{i,j}$, and hence in a "rectangular" submodule $$\bigoplus_{0\le i,j\le n} M_{i,j}.$$ Any element here can be "pushed" to one supported only in $M_{n,n}$, and it is easy to check what it means that it vanishes there. The same argument generalizes to higher "dimensions" $$\bigoplus M_{i,j,k}.$$
You want to take $$B:= A\left[\{x_s\}_{s\in S}\right]$$ and consider some ideal generated by elements of the form $(sx_s - 1)$. Considering the polynomial ring as a free $A$-module, the required statement follows from the previous one, once one shows the ideal is generated as an $A$-submodule of $B$ by elements of the same form as above. This is not so bad since $$(sx_s -1)(tx_t - 1) = (stx_sx_t - 1) - (sx_s -1) - (tx_t - 1)$$ is already a sum of three relations, and the same will occur for more complicated monomials in place of $x_s$ and $x_t$. Multiplying a relation by a monomial also gives you a sum of relations in the quotient submodule of $$\bigoplus_{i,j}M_i,j$$ above.
