For which logarithm bases is this sum finite?

For base 2, you have a sum of 1's which diverges. For smaller bases, use that for every recursion,

$$ \log_b(x) = \log_2(x) \cdot \log_b(2) > \log_2(x) $$

so the terms in your sum get even larger.

So for $b<2$, the sum diverges.

If it is known (as stated by the author) that the series diverges for $b=e$, then by the same argument it will also diverge for $b<e$.

It remains to prove that for $b> e$, the sum converges. We will use the well known fact (for base e) that $\log(1+x) \leq x$. We have

$$ a_{n+1} = \log_b(1 + a_n) = \log_e(1 + a_n) \cdot \log_b(e) \leq a_n \cdot \log_b(e) $$

so

$$ \sum_{n=0}^\infty a_n \leq \sum_{n=0}^\infty (\log_b e )^n $$

which converges (geometric series) exactly for $\log_b e <1$ or $b > e$.

Hence, in total, the series converges for $b > e$.