I think it might depend on the choice of points. Here is one example of a knotless embedding. Let the points on $A$ be $\{(-4,0,0), (-1,0,0), (1,0,0), (4,0,0)\}$. Let the points on $B$ be $\{(0,-3,1), (0,-2,1), (0,2,1), (0,3,1)\}$. Look at the projection $\pi$ of $K_{4,4}$ onto the $xy$ plane, which has exactly four crossings. A non-trivial knot has at least three crossings in any diagram.
Suppose $L$ is a non-trivial knot or link embedded in this $K_{4,4}$. Then $L$ has at most 8 edges. If $\pi(L)$ contains all four crossings, then $L$ is a pair of unlinked circles. If $\pi(L)$ contains three crossings, then by symmetry, it doesn't matter which three are chosen, and $L$ is an unknot. Therefore this embedding is knotless. There is however a Hopf link in this $K_{4,4}$, which uses two opposite crossings, so this embedding is not linkless.
Different choices of points could give many more crossings in the projection, and I wouldn't like to try to extend this argument to anything much more complicated.