Seems to be an easy question containing lots of cases: $5$ points are on one circle $\Gamma$ $A,B,C,D,E$ then other points have to be also on this circle. $4$ points are on a circle, then we can do yet $AB \cap CD, AD \cap BC, AC \cap DB$ as fifth point. Now each $3$ of $4$ points have to be on a line $l$ and so we have max. $1$ point not on that line.

$S$ is a circle or a line. If $S$ have three point on a line then $S$ is a line. If $S$ have no some three point on a line then $S$ is a circle

If $S$ has size $0, 1, 2$, or $3$ it satisfies the condition vacuously, so consider $S$ with $|S| \ge 4$; we'll use the notation $(WXYZ)$ to denote the result of applying the condition to points $W, X, Y, Z$. Suppose there are four points $A, B, C, D \in S$ which are concyclic. If there is a fifth point $E$ on the same circle, all points must lie on this circle; to see this, consider any other point $F$ not on the circle. $(FABC)$ forces, WLOG, $F \in AB$. $(FCDE)$ forces, WLOG, $F \in CD$. Finally, $(FBDE)$ gives $F \in BD$ or, WLOG, $F \in BE$. In the first case, $F \in AB$ and $F \in BD$ so $F \equiv B$; in the second, $F \in AB$ and $F \in BE$ so $F \equiv B$. Either way, contradiction. Instead if $A, B, C, D$ are concyclic and the only four points of $S$ on their circumcircle, consider a fifth point $E$. $(ABCE)$, $(ABDE)$, $(ACDE)$, $(BCDE)$ force $E$ to be the intersection of the two lines joining two pairs of the points $\{A, B, C, D\}$; that is $E$ is one of $AB \cap CD$, $AC \cap BD$, $AD \cap BC$. Any of these work and prohibit a sixth point $F$; WLOG suppose $E = AB \cap CD$ so that $F$ is either $AC \cap BD$ or $AD \cap BC$. In the first case, $(FADE)$ yields (using the fact that $E, F\not \in AD$) either $E, F, A$ or $E, F, D$ collinear. Yet $E$ and $F$ lie on different lines through $A, D$, contradiction. The second case follows analogously, this time looking at $(FACE)$. Finally, suppose no four points of $S$ are concyclic. In particular, since $|S| \ge 4$, some three of them $A, B, C$ are collinear. When there are exactly $0$ or $1$ points in $S$ not on this line, the condition is clearly satisfied; if there are two or more ($D$ and $E$), then $(ABDE)$ forces, WLOG, either $D \in AB$ or $A \in DE$. The first case contradicts the fact that $D$ is not on line $ABC$, so $A \in DE$ and $A \in ABC$ meaning $A = ABC \cap DE$. Now $(BCDE)$ gives the desired contradiction. In conclusion, $S$ can be any collection of points on a single circle, any set of four concyclic points $A, B, C, D$ together with exactly one of the points $AB \cap CD, AC \cap BD, AD \cap BC$, or any collection of points from a single line together with at most one point not on that line (note any set $S$ of size at most $3$ is covered here as well).