With $2s-4$ points, divide them into two groups of $s-2$ points each. Edges within a group are red, edges between groups are blue.

You have established $2s-2$ points must have a star. Here I show $2s-4$ points don't have to have a star. That still leaves $2s-3$ points...

With $2s-4$ points, divide them into two groups $A$ and $B$, of $s-2$ points each. Edges between two $A$ points are red; edges between two $B$ points are red; and edges between an $A$ point and a $B$ point are blue. Each point is attached by $s-3$ red edges and $s-2$ blue edges, so we don't get an $s$-star.