Below we prove by induction that
$$c = \frac{1 + \sqrt{1 + \frac{4}{d}}}{2},$$
which is the positive root of the equation $c^2 - c - 1/d = 0$.

First we prove that the statement holds for $n = 0$. By definition $b_0 = a_0$, and
$$\begin{aligned} [ca_0] &= \left[ \frac{1 + \sqrt{1 + 4/d}}{2} a_0 \right] = a_0 + \left[ \frac{-1 + \sqrt{1 + 4/d}}{2} a_0 \right] \\ &= a_0 + \left[ \frac{2}{1 + \sqrt{1 + 4/d}} \cdot \frac{a_0}{d} \right] = a_0, \end{aligned}$$
so $b_0 = [ca_0]$ holds.

Next we prove by induction that the statement holds for all $n > 0$. By the problem statement, the value of $b_n$ is obtained from $a_n$ by adding 1 some number of times: when performing the operation $S_k \rightarrow S_{k+1}$, if that term is increased by 1, then necessarily $b_k + k \ge n$ and $0 \le k < n$. For every $n$, we write the set of such $k$ as
$$X_n = \{k \mid b_k + k \ge n \text{ and } 0 \le k < n\},$$
and then we have $b_n = a_n + |X_n|$. The key lies in how to compute the size of the set $X_n$: by induction, we have
$$\begin{align*} X_n &= \{k \mid [ca_k] + k \ge n \text{ and } 0 \le k < n\} \\ &= \{k \mid ca_k + k \ge n \text{ and } 0 \le k < n\} \\ &= \{k \mid ca_0 + k(1 + cd) \ge n \text{ and } 0 \le k < n\} \\ &= \{k \mid \frac{n - ca_0}{1 + cd} \le k < n\} \end{align*}$$
Note: here it does not matter even if $n < ca_0$, because
$$\frac{n - ca_0}{1 + cd} > -1$$
holds always, and this does not affect the count of $k$.

Therefore
$$\begin{align*} b_n &= a_n + |X_n| = a_n + \left[ \frac{ca_n}{1+cd} \right] \\ &= \left[ \left(1 + \frac{c}{1+cd}\right) \cdot a_n \right] = [ca_n], \end{align*}$$
and the original statement is proved.