Note that actually, $\cap_{i\in \emptyset} A_i = S$ is not a convention, it's actually just wrong.
The correct statement would be $\cap_{i\in \emptyset} A_i = \emptyset$, or $\bigwedge_{i\in \emptyset} A_i = S$, where $\bigwedge$ denotes the greatest lower bound in the complete lattice $(\mathcal{P}(S), \subset)$.
Why are these true ? Well first note that $\cap_{i\in \emptyset} A_i = S$ simply cannot be true. One the LHS we have an intersection of sets with no mention of $S$ (the $A_i$'s are subsets of $S$, but they're also subsets of $S\cup \{S\}$ or god knows what), and on the RHS we have a set $S$, that happens to contain the $A_i$'s. So thats just wrong because then, $\cap_{i\in \emptyset} A_i = S\cup\{S\}$ would also be true (the $A_i$s are subsets of $S\cup\{S\}$, after all), which is absurd.
So $\bigcap_{i\in I}A_i$ is actually an abbreviation for $\bigcap \{A_i \mid i\in I\}$ which is by definition $\{x\in \displaystyle\bigcup_{i\in I} A_i \mid \forall i\in I, x\in A_i\}$, where $\displaystyle\bigcup_{i\in I}A_i$ is an abbreviation for $\bigcup\{A_i\mid i\in I\}$ which is $\{x\mid \exists i\in I, x\in A_i\}$ (its existence is provided by the axiom of reunion). But then if $I=\emptyset$, this last thing is empty, and so are all its subsets, in particular $\cap_{i\in \emptyset} A_i $, which must then be empty.
However, in the aforementioned complette lattice, $\cap$ and $\wedge$ coincide on all sets but the emptyset. It is easy to see that in a complete lattice, the greatest lower bound of the emptyset is the lattice's maximum element. Therefore $\bigwedge_{i\in \emptyset}A_i = \bigwedge\emptyset = S$. Note that here, $\bigwedge$ does depend on $S$ so this makes complete sense.
However, in practice, since $\bigwedge$ and $\cap$ coincide so often, we ignore those subtleties and declare them to be equal, which leads to the abuse of notation $\bigcap_{i\in \emptyset}A_i = S$, which is what the other answers essentially tell you. But this is an abuse of notation (by the usual definitions etc.)