I'm not sure if this is a well researched problem, I'm trying to think of a solution, it does not seem very obvious.
For eg: If $n = 6$ and $s = 3$, we can introduce everyone in $4$ iterations
1) A B C | D E F
2) A E F | D B C
3) A D E | F B C
4) A D F | E B C
I don't want to figure out a scheme of grouping manually for $n = 12$ and $s = 3$, please point me to relevant discussion or literature, if someone identifies what problem it is.
Let $G=(V,E)$ be a graph with $n$ vertices (the number of people).
You defined an operation, that takes $2s$ vertices, and connects them by edges (disregarding the current edges between them) to construct the graph $K_{s,s}$ (Complete bipartite graph).
You asked whats the minimal number of rounds required to construct the graph $K_n$.
Since $\vert E \vert = {n \choose 2}$, and you add at most $s^2$ edges each time, a lower bound on the number of rounds you must perform is ceil of $\frac{{n \choose 2}}{s^2}$
It wasn't clear to me if you are looking for an number or an algorithm to find a solution, so here is an algorithm:
I'm not sure if a greedy algorithm will obtain the optimal result, but it will obtain $O(ln(s))$ approximation. (multiplicative factor)
Great question!