Let $S$ be a multiplicative subset of a ring $A$. I have just written down a proof for $\varinjlim\limits_{s\in S} A_s = S^{-1}A$.
For this I defined a preorder on $S$ by setting
$s\geq t$ if and only if there exists some $u$ $\textbf{in S}$ such that $us=t$
and took the corresponding thin category as an index set for the obvious diagram and then checked the universal property of the colimit which seemed all pretty straight forward. But when reading through the proof, I was just wondering why I didn't define the preorder slightly differently:
$s\geq t$ if and only if there exists some $u$ $\textbf{in A}$ such that $us=t$,
which should lead to a bigger diagram than the above.
I can't figure out where this should make a difference in the proof. So it seems to me that $S^{-1}A$ is the colimit of both diagrams. For constructing a morphism $S^{-1}A \to N$ in the universal property of the colimit, it is only required that, for each triple $s,t,u$ $\textbf{in S}$, the triangles $(A_s,A_{stu},N)$ and $(A_t,A_{stu},N)$ commute, which suggests that the first definition of the preorder is sufficient.
If I am right, I would appreciate an intuitive explanation for why both colimits agree.
I would start pointing out that probably you wanted to consider the opposite preorder (the one where $s\leq t$ if and only if $t = su$), otherwise is not so clear to me what should be the morphism $A_s \to A_t$.
The two colimit coincide, the reason is due to the fact that if $D \colon P \to \mathbf C$ and $D' \colon P' \to \mathbf C$ are two diagrams, over the same category, and $H \colon P \to P'$ cofinal functor, then the colimit of $D$ and $D'$ are isomorphic.
A proof of this fact and the relative definition of cofinal functor can be found in chapter 0 of Adamek et al.'s Locally presentable and accessible categories.
This theorem can be applied to the case at hand by letting $P$ and $P'$ being the two preorders, $D$ and $D'$ the two diagrams of localized rings and, finally, $H$ the embedding between the two preorders (which is cofinal because is surjective between filtered categories).
If you need additional detalis feel free to ask in the comments below.