Suppose otherwise. Let the period of $a_0,...,$ be $t$, and $\exists c$ such that infinitely many such $n$ are congruent to $c \pmod p$, and $a_c$ is part of a repeating block. Let those $n$ be $c + m_it \ (i=0,1,...)$ and $m_0 = 0$. Then $c+m_it | Ab^{m_it} + \frac{s(b^{m_it}-1)}{b^t - 1}$ where $s$ is the numerical value of the repeating block. Write $c+m_it = \prod_j p_j^{\alpha_j}$ we deduce that $$\forall \ j, \ Ab^{\prod_j p_j^{\alpha_j} -c} + \frac{s(b^{\prod_j p_j^{\alpha_j}-c} -1)}{b^t-1}\equiv 0 \pmod {p_i^{\alpha_i}}$$ First note that if $p_i | b$ then $p_i^{\alpha_i}| \frac{s(b^{\prod_j p_j^{\alpha_j}-c}-1)}{b^t - 1}$ but its $p$-adic valuation is $\nu_p(s)$, a contradiction for $\alpha_i$ large. Thus for large enough $\alpha_i > \nu_p(s) + 1$ we must have $(b,p_i) = 1$. Write $\nu_{p_i}(s) = \sigma,p = p_i$ with $\alpha := \alpha_i -\sigma > 1$ and we obtain by letting $k = \frac{(A+\frac{s}{b^t-1})}{b^cs}$ that $$kb^{\prod_j p_j^{\alpha_j}} - 1 \equiv 0 \pmod {p^\alpha}.$$ Let $l = \prod_j p_j^{\alpha_j}$ it follows that $p^\alpha | kb^l -1.$ For each $n \in S$ define the peak primes set of $n$ as the set of primes $p$ with $\nu_p(n) > 1 + \nu_p(s)$. For all $n$ sufficiently large, the peak prime set is nonempty, and there are infinitely many such $n$. Since the set of prime divisors of $S$ is finite, it follows that infinitely many $n$ have the same nonempty peak prime set. Within these, we define a majorization relation $x \succ y$ to mean $\forall p | S, \nu_p(x) > \nu_p(y)$ (which is essentially the same as $y|x$). From here it is easy to see that there must be an infinite chain $x_0 \prec x_1 \prec x_2 ...$. It follows that $kb^{x_0} -1 = p^\beta r$ for some $\beta \ge \alpha, p\nmid r$. Letting $x_i =mx_0$ with $p^\mu || m$ to get $kb^{m x_0} - 1 \equiv 0 \pmod {p^{\mu + \alpha}}$. However $p^{\beta + \mu} | k^m b^{m x_0} - 1$, hence $p^{\mu + \alpha} | k^{m - 1} - 1 = (k^m b^{m x_0} - 1) -k^{m - 1} (kb^{m x_0} -1)$. Let $m_i = \frac{x_i}{x_0}$ now to get $p| k^{m_i - 1} - 1$ for all $m_i$, and the $\gcd$ of those is $k^g - 1:=k^{\gcd _{i\ge 1}(m_i - 1)} - 1 | k^{m_0}-1$, so $\nu_p(k^g -1)$ is positive and $< \mu+ \alpha$ for all $i$ sufficiently large. From here on we conlude that $p|\frac{m_i - 1}{g} | m_i - 1$, a contradiction to $p|m_i$.

Denote $\frac{x}{y}:=\sum_{j=1}^{\infty} \frac{a_i}{b^i}$ By base conversion rules $(a_0a_1\dots a_n)_b =\left\lfloor \frac{xb^n}{y}\right \rfloor$ Let $(a_0a_1\dots a_n)_b =\frac{xb^n-t}{y}$ Since for all $n \in \mathbf{S},n|(a_0a_1\dots a_n)_b =\frac{xb^n-t}{y}$ we must have $xb^n \equiv t \pmod {yn}$ Claim: Let $p$ be any prime such that $\nu_p(n)>1$.Then $\nu_p(xb^n-t)$ is eventually constant. Proof: FTSOC,assume that $\nu_p(xb^n-t)$ isn't eventually constant. Let $n=mp^k$ with $\gcd (m,p)=1$ and let $F$ be an integer such that $\frac{b^F}{x} \equiv 1 \pmod p$(which exists) Then choose $X=mp^k(p-1)F\in \mathbf{S}$ to get $1 \equiv x^{p-1} \pmod {p^k}$ for all sufficiently large $k$, contradiction so we are done.(We get all $x,b,y,n=1$ so $a_i=0$ which isn't true) By the claim we are done.