We have
$$
\begin{align*}
a\left(1-\frac{1}{b}\right)^{a-1} &= r \\
a\left(1-\frac{1}{b}\right)^a &= \left(1-\frac{1}{b}\right)r \\
a e^{a \log\left(1-\frac{1}{b}\right)} &= \left(1-\frac{1}{b}\right)r \\
a \log\left(1-\frac{1}{b}\right) e^{a \log\left(1-\frac{1}{b}\right)} &= \left(1-\frac{1}{b}\right)r\log\left(1-\frac{1}{b}\right),
\end{align*}
$$
so that
$$
\begin{align*}
a \log\left(1-\frac{1}{b}\right) &= W\left(\left(1-\frac{1}{b}\right) r \log\left(1-\frac{1}{b}\right)\right) \\
a &= \frac{W\left(\left(1-\frac{1}{b}\right) r \log\left(1-\frac{1}{b}\right)\right)}{\log\left(1-\frac{1}{b}\right)},
\end{align*}
$$
where $W$ is the the Lambert W function. Note that $W(x)$ is double-valued when $x \in (-1/e,0)$, and the solution you want is given by the principal branch:
$$
a_0 = \frac{W_0\left(\left(1-\frac{1}{b}\right) r \log\left(1-\frac{1}{b}\right)\right)}{\log\left(1-\frac{1}{b}\right)}
$$
If you'd like, you can expand this in a series which converges for $b$ large:
$$
a_0 = r + \frac{r(r-1)}{b} - \frac{3r^2(r-1)}{2b^2} + O(b^{-3}).
$$
Let us denote the other solution, given by the other branch of $W$, by
$$
a_{-1} = \frac{W_{-1}\left(\left(1-\frac{1}{b}\right) r \log\left(1-\frac{1}{b}\right)\right)}{\log\left(1-\frac{1}{b}\right)}.
$$
One can use the asymptotic series derived in this paper this paper (pp. 19-23) to calculate an expression for this solution as $b \to \infty$:
$$
a_{-1} = -\frac{1}{\log\!\left(1-\frac{1}{b}\right)}\left\{\log b + \log \log b - \log r + \frac{\log \log b}{\log b} - \frac{\log r}{\log b} + O\!\left(\frac{\log \log b}{\log b}\right)^2\right\}.
$$
This is an okay approximation but note that the absolute error does not decrease to $0$ so it won't be very helpful for numerics.