this question must be the most stupid I have ever asked.
If $A$ is a commutative ring and $S$ a multiplicative subset the usual inclusion induces a homeomorphism onto the image $\text{Spec} (S^{-1}A) \rightarrow \text{Spec} (A)$ and, then, it's a mono.
By the anti-equivalence between affine schemes and commutative rings, the inclusion $A \rightarrow S^{-1} A$ must be an epi. However this would be equivalent to the statement that given $y \in A$ and $s \in S$ there exists $x \in A$ and $s' \in S$ such that $s'xs = s'y$ , which apparently is not true for any ring $A$.
So what's the problem? And if it's true, how to prove the above statement without the anti-equivalence?
No for "this would be equivalent to ...". See MO/109 for some results about epimorphisms of commutative rings. (Don't confuse them with regular epimorphisms of commutative rings, which are surjective.)
Also, it is not true that a morphism of schemes which is a homeomorphism onto its image is a monomorphism. Certainly it becomes a monomorphism of topological spaces, but applying the forgetful functor is often not enough! It is true that $A \to S^{-1} A$ is an epimorphism of commutative rings (by the universal property) and hence $\mathrm{Spec}(S^{-1} A) \to \mathrm{Spec}(A)$ is a monomorphism of affine schemes. Actually, it is a monomorphism in the whole category of schemes because $\mathrm{Spec} : \mathsf{CRing}^{op} \to \mathsf{Sch}$ is right adjoint to $\Gamma$ and hence preserves monomorphisms.