Is there a nonempty planar set that contains $0$ or $2$ vertices from each unit equilateral triangle?

I know that such a set cannot be measurable. In fact, my motivation is to extend a Falconer-Croft proof that works for measurable sets, see the details here. In more general, to make their proof work for other sets besides equilateral triangles, one can ask the following.

Suppose we are given two sets, $S$ and $A$ in the plane, such that $S$ is finite, with a special point, $s_0$, while neither $A$ nor its complement is a null-set, i.e., the outer Lebesgue measure of $A$ and $A^c=\mathbb R^2\setminus A$ are both non-zero. Can we find two congruent copies of $S$, $f_1(S)$ and $f_2(S)$, such that $f_1^{-1}(f_1(S)\cap A)\Delta f_2^{-1}(f_2(S)\cap A)=\{s_0\}$, i.e., $s_0$ is the only element of $S$ that goes in to/out of $A$ when we go from $S_1$ to $S_2$?