Let $A$ be a commutative ring with $1$ and let $S\subset A$ be a multiplicative set such that $1\in A$.
In the usual construction of the localization $S^{-1}A$, we have $$\ker{l}=\{a\in A:\exists s\in S(sa=0)\},$$ where $l\colon A\to S^{-1}A$ maps $a\in A$ to $\frac{a}{1}\in S^{-1}A$, essentially because $S^{-1}A=A\times S/\sim$, where $(a,s)\sim (b,t)$ iff there exists $u\in S$ such that $u(at-bs)=0$.
Is it possible to prove the above equality just from the universal property of $l\colon A\to S^{-1}A$ without recurring to any particular construction of $S^{-1}A$?
The fact that $l(s)\in S^{-1}A$ is invertible for all $s\in S$ implies $\{a\in A:\exists s\in S(sa=0)\}\subset\ker{l}$. The problem is the reverse inclusion.
You can't really prove a negative like this, but I feel like the answer has to be basically no. In order to use the universal property, you have to actually have an example of a ring-homomorphism $f:A\to B$ that sends every element of $S$ to a unit. In particular, if you want to restrict the kernel of $l$, you would need to give, for each $a\in A$ that does not annihilate any element of $S$, some particular ring-homomorphism $f:A\to B$ such that $f(a)\neq 0$ and $f(s)$ is a unit for each $s\in S$.
So no matter what, in order to use the universal property, you need to construct some particular rings $B$. You technically don't have to choose $B$ to be the localization itself, but it is hard for me to imagine any choice of $B$ that would be more natural, or how you could construct such a $B$ in full generality without doing basically the same work you have to do to construct the localization.