The points should be on a circle (or line) and $A$, $B$ should separate $A'$, $B'$ on it. To prove necessity, first suppose that the points are not coplanar. Then there exist two parallel planes passing through $A$, $B$ and $A'$, $B'$ respectively. Any two circles in these planes are not linking. So the points should be coplanar.

Now suppose $B'$ is not on the circumcircle of $ABA'$ (which can be a line). So we can slightly change the circle to find a circle passing through $A$, $B$ such that $A'$, $B'$ are both outside or both inside it. Now, this circle is not linking with the circle with diameter $A'B'$ orthogonal to the plane containing the points.

So the points should be on a circle (or line). Now, suppose $A$, $B$ do not separate $A'$, $B'$ on the circle. If we change the circle slightly, still passing through $A$, $B$, then $A'$, $B'$ will be both inside or both outside the new circle and we arrive to a contradiction like the previous case. So, the necessity of the condition is proved.

To prove sufficiency, Let $C$, $C'$ be two different circles passing through $A$, $B$ and $A'$, $B'$ respectively. Let $P$, $P'$ be the planes containing $C$, $C'$ respectively. If the points are collinear, then $C' \cap P$ is consisted of a point inside $C$ and a point outside $C$. So $C$, $C'$ are interlocked. So, suppose the points are on a circle. Let $M$ be the intersection of the segments $AB$ and $A'B'$. We have $MA \cdot MB = MA' \cdot MB'$. Let

$l = P \cap P'$ which passes through $M$. $M$ is inside $C$, so $l$ intersects $C$ at two points like $X,Y$ and $M$ is between $X,Y$. Similarly, $l$ intersects $C'$ at $X',Y'$ namely, and $M$ is between $X',Y'$. Suppose $X,X'$ are in one side of $M$. We have
$$MX \cdot MY = MA \cdot MB = MA' \cdot MB' = MX' \cdot MY'$$
So if $MX \le MX'$, then $MY \ge MY'$ and vice versa. So the points of $C' \cap P = \{X',Y'\}$ are in different sides of $C$ or both are on $C$. So $C,C'$ are linking and sufficiency of the condition is proved. $\square$