I believe the answer to Question 1 is positive, though maybe I've misunderstood something and the following sketch misses something important.

For $n \in [1,N]$, the DFA halts and decides whether $n \in S$ given the $b$-ary expansion of $n$, which has length $\lfloor \log_b N \rfloor$, left-padded with $0$'s if necessary.

As OP says, iid input to a DFA yields a Markov chain on the underlying (directed) graph. The probability of not reaching a sink after seeing the $T$th digit is $O(\lambda^T)$ where $0 < \lambda < 1$ is the Perron eigenvalue of a matrix related to the transition matrix for the original Markov chain. Let $p$ be the limiting probability of accepting.

Compare the density $\delta_N(S) = \frac{1}{N}|S \cap [1,N]|$ to the probability of accepting after $T$ symbols where $\lfloor \log_b N \rfloor \leq T \leq \lceil \log_b N \rceil$. The probability of not reaching a sink after seeing the $T$th digit is then $O(N^{-\alpha})$ where $\alpha = \log_b(1/\lambda) > 0$. So you have $\delta_N(S) - p  = O(N^{-\alpha})$.