It is known that the following are equivalent for an epimorphism $A \to B$ in $\mathbf{CRing}$:

Let $S$ be the set of elements $a \in A$ such that $A [a^{-1}] \to B [a^{-1}]$ is an isomorphism. Then $S$ generates the unit ideal in $B$.

$A \to B$ is a flat homomorphism of finite presentation.

$A \to B$ is an étale homomorphism.

What I'd like to know is if there is an *algebraic* proof that 2 implies 1.

Here is a cheat proof: it is well known that flat morphisms locally of finite presentation are open maps, so $\operatorname{Spec} B → \operatorname{Spec} A$ factors as an fppf monomorphism $\operatorname{Spec} B → U$ followed by an open immersion $U → \operatorname{Spec} B$; but fppf monomorphisms are isomorphisms, so we are done. This proof is unsatisfactory to me because (a) it involves a scheme that is not known to be affine *a priori* and (b) the well-known fact about flat morphisms locally of finite presentation being open is (to me, at least) a deep result in scheme theory.

(Incidentally, is there a good name for ring homomorphisms corresponding to open immersions? I want to say "open localisation", but that's misleading because they are not necessarily localisations...)