This year I submitted my work on this problem as part of Summer of Math Exposition
https://btm.qva.mybluehost.me/solving-freudenthals-impossible-problem/

This includes a much better result than my 2018 comment of a hyperexponential sequence with $F(s)$ at most 1 (but likely 0).

In the arithmetic progressions $6705290747541849491 + (2^{65}-2)n$ and $30188197382697384551+(2^{65}-2)n$, there exists a proven alternate factor sum in $S$ for every pair of numbers except $(m, 2^km)$ and $(a,2^k)$.

Using this, I was able to conclude that the expected value of $F(s)$ on these progressions goes as $\log(s)^{-2}$, and because the progressions are themselves linear, this *strongly* suggests that there are infinitely many $s$ with $F(s) =1$ on this progression.

For a (rather extreme) example, I believe I have shown that $F(3497072530534972582665793*(2^{4096}+1)) = 1$, leading to the new largest known solution $(3497072530534972582665793, 3497072530534972582665793*2^{4096})$.