ac.commutative algebra - equality of elements in localization via universal property [unsolved!]

I'm not sure I understand the question. More exactly, I think I disagree with what seems to be an assumption built into it: that there are sharp lines to be drawn between 'only using the universal property' and not, or between working 'without elements' and not.

Generally, a universal property describes how something interacts with the world around it. For example, if you say that an object $I$ of a category $mathcal{C}$ is initial, that describes how $I$ interacts with other objects of $mathcal{C}$. If you don't know much about the objects of $mathcal{C}$, it doesn't tell you much about $I$. Similarly, you're not going to be able to deduce anything about the ring $S^{-1}A$ without using facts about rings. I don't know which of those facts are ones you'd be happy to use, and which aren't.

There's nothing uncategorical about elements. For example, you're dealing with rings, and an element of a ring $A$ is simply a homomorphism $mathbb{Z}[x] to A$. (And $mathbb{Z}[x]$ can be characterized as the free ring on one generator, that is, the result of taking the left adjoint to the forgetful functor $mathbf{Ring} to mathbf{Set}$ and applying it to the terminal set 1.)

So I'm unsure what exactly your task is. But I'd like to suggest a different universal property of localization, which might perhaps make your task easier. Here it is.

Let $mathbf{Set}/mathbf{Ring}$ be the category of rings equipped with a set-indexed family of elements. Formally, it's a 'comma category'. An object is a triple $(S, i, A)$ where $S$ is a set, $A$ is a ring, and $i$ is a function from $S$ to the underlying set of $A$. (You might think of $i$ as an including $S$ as a subset of $A$, but $i$ doesn't have to be injective.) A map $(S, i, A) to (S', i', A')$ is a pair $(p, phi)$ consisting of a function $p: S to S'$ and a homomorphism $phi: A to A'$ making the evident square commute.

There is a functor $R: mathbf{Ring} to mathbf{Set}/mathbf{Ring}$ given by
$$
R(A) = (A^times to A)
$$
where $A^times$ is the set of units of $A$ and the arrow is the inclusion. Then $R$ has a left adjoint $L$, given by $L(S, i, A) = (iS)^{-1}A$. In other words, the left adjoint to $R$ is localization.