I think the probabilistic method gives an A of size $O(n^{d/2}d\log n)$.

Put every point to A independently with probability p. What is the probability that a point x will be in S(A)? For any x, we can find $\Omega(n^d)$ pairs of points that are all different such that x lies on the line of any pair. (This is true because e.g. for d=2 if x is in the bottom-left part of $K_n$, then we can take the $n/4\times n/4$ grid upper-right from it, contract the $n/8 \times n/8$ bottom-left of this grid, and double each point from x to get its pair.)

The probability that both points of a fixed pair are in A is $p^2$, the probability that no such pair exists is $(1-p^2)^{n^d}$. So if $n^d(1-p^2)^{n^d}<1$, then we are done using the union bound. Unless I am mistaken this is true if $p>\Omega(n^{-d/2}d\log n)$.

Now of course we cannot be sure about how big A is. But if we replace the above $<1$ with a $<1/2$, then we can even add the condition that A should be at most $pn^d$, for which the probability is $\ge 1/2$. So we get $O(n^{d/2}d\log n)$ points. Maybe this can be further improved with some more advanced probabilistic methods.

I think the probabilistic method gives an A of size $O_d(n^{d/2}\sqrt\log n)$. (Formula updated according to js's comment.)

Put every point to A independently with probability p. What is the probability that a point x will be in S(A)? For any x, we can find $\Omega(n^d)$ pairs of points that are all different such that x lies on the line of any pair. (This is true because e.g. for d=2 if x is in the bottom-left part of $K_n$, then we can take the $n/4\times n/4$ grid upper-right from it, contract the $n/8 \times n/8$ bottom-left of this grid, and double each point from x to get its pair.)

The probability that both points of a fixed pair are in A is $p^2$, the probability that no such pair exists is $(1-p^2)^{n^d}$. So if $n^d(1-p^2)^{n^d}<1$, then we are done using the union bound. Unless I am mistaken this is true if $p>\Omega_d(n^{-d/2}\sqrt\log n)$.

Now of course we cannot be sure about how big A is. But if we replace the above $<1$ with a $<1/2$, then we can even add the condition that A should be at most $pn^d$, for which the probability is $\ge 1/2$. So we get $O_d(n^{d/2}\sqrt\log n)$ points. Maybe this can be further improved with some more advanced probabilistic methods.