Let $s$ consist of $2^{2^k}$ zeros, followed by the same number of ones, for increasing $k$:
$$01\, 0^21^2\, 0^{16}1^{16}\, 0^{256}1^{256}\,0^{65536}1^{65536}\dots$$
Then $s\circ f_{a,b}$ will have all (but one) blocks of 1s of sizes $\lfloor2^{2^{k}}/a\rfloor$ and $\lceil 2^{2^k}/a\rceil$.


If $2^{2^k}/{a_1}\le 2^{2^j}/{a_2}+1$ then $2^{2^k-2^j}\le a_1/{a_2}+2^{-2^j}$.
But if $k>j\to\infty$ then $2^{2^k-2^j}$ is unbounded.

(Note that $2^k/4=2^{k-1}/2$, so $2^k$ in place of $2^{2^k}$ is not enough for this construction.)

In conclusion, all $s\circ f_{a,b}$ are distinct.