A question about the equivalence relation on the localization of a ring. Let $A$ be a ring and $S$ a multiplicative closed set. Then the localization of $A$ with respect to $S$ is defined as the set $S^{-1}A$ consisting of equivalence classes of pairs $(a, s)$ where to such pairs $(a,s), (b,t)$ are said to be equivalent if there exists some $u$ in $S$ such that 
$$u(at-bs)=0$$
Now, in the Wikipedia article about the localization of a ring, it says that the existence of that $u\in S$ is crucial in order to guarantee the transitive property of the equivalence relation. 
I've seen the proof that the equivalence relation defined above is indeed an equivalence relation, but I fail to see how crucial the existence of $u$ is. For example, why doesn't it work if we simply say that two pairs $(a,s),(b,t)$ are equivalent iff $at - bs = 0$? I tried to come up with a counterexample for such case, but failed in the attempt. 

For a more geometric example, let $A = \mathbb{R}[x,y] / (xy)$ and $S = \{ 1, x, x^2, x^3, \ldots \}$.  (In algebraic geometry, $A$ represents a ring of functions on the union of the two axes in $\mathbb{R}^2$.)  Then in $S^{-1} A$, $y/1 = 0/1$ even though $y \cdot 1 - 0 \cdot 1 \ne 0$ in $A$ - but in fact, $x (y \cdot 1 - 0 \cdot 1) = 0$.
Intuitively, the reason $y = 0$ must hold in $S^{-1} A$ is that $xy = 0$ is inherited from $A$, and then you can multiply both sides by $x^{-1}$.  (The localization corresponds to the geometric operation of intersecting with the set where $x \ne 0$.)
