2 approaches:
Generate $b$ randomly, having $b(n)$ be $0$ with probability $1/2$ and having each $b(n)$ be independent. Then for each $(a,b)$ and $(a_1,b_1)$, the probability that $$ b\circ f_{a,b}=b\circ f_{a_1,b_1} $$ is $0$. Summing over all countably many such pairs, we still get a probability of $0$, so with probability $1$, our string works.
Generate $b$ greedily. Go through pairs $(a,b),(a_1,b_1)$ in some order and for each, choose two entries in our string very far out to force disagreement.