Set $b_i = 2\ell_i$, all of them are even. At the first step we get the pairs
$$A := \{(p, q) : (p, q) = (1, 2\ell_1), \ell_1 \in \mathbb{Z}. \quad (1)\}.$$
At each subsequent step we expand the set $A$ by adding additional pairs $(p', q')$
defined as
$$\left\{ (p', q') : \frac{p'}{q'} = \frac{1}{2\ell + \frac{p}{q}}, (p, q) \in A, \ell \in \mathbb{Z} \right\} \quad (2).$$
By (2) we obtain
$$p' = q; \quad q' = 2\ell q + p \quad (3)$$
So, if $(p, q)$ is obtained in the process of expanding, we also add $(p', q')$ defined as in (3). We want to characterize all pairs $(p, q)$ that can be obtained in this way. Note that if $(p, q) = 1$, then from (3) follows $(p', q') = 1$. Let us prove that the set of pairs $(p, q)$ in question is
$$A := \{(p, q) : p, q \in \mathbb{Z} \setminus \{0\}, |p| < |q|, (p, q) = 1, \text{ one of } p, q \text{ is even, the other is odd.}\}$$
Note that we ruled out $|p| = |q|$ because $\frac{p}{q} = \pm 1$ cannot be represented as wanted. The transformation described in (3) shows that we cannot step outside $A$. To prove that all pairs in $A$ can be generated, we follow a standard procedure - take $(p', q') \in A$ and search for $(p, q) \in A$ that generates $(p', q')$ via (3), but we want $(p, q)$ be "less" than $(p', q')$. That's how induction works. Solving (3) with respect to $(p, q)$ yields
$$p = q' - 2\ell p'; \quad q = p' \quad (4)$$
Clearly, we can choose $\ell \in \mathbb{Z}$ such that $|q' - 2\ell p'| < |p'|$ (since $p' \neq 0$). Indeed, consider the points $q' - 2\ell p'$ where $\ell$ runs through all integers. These points are at a distance $2p'$ apart, and none of them hits $p'$ because $(p', q') = 1$. Hence, the point closest to 0 has magnitude less than $p'$. So, we started from $(p', q') \in A$, and found $(p, q) \in A$ that generates $(p', q')$ and moreover, $q = p', |p| < |p'| < |q'|$. Apparently $(p, q)$ is "less" than $(p', q')$, that is, $|p| < |p'|, |q| < |q'|$. Now, induction does the job. Obviously, $(1, 2), (-1, 2), (1, -2), (-1, -2)$ can be represented as continued fractions satisfying the statement. Assume that all $(p, q) \in A$ with $|p| \le N, |q| \le N$ can be represented like that. Take $(p', q') \in A, |q'| = N + 1$. As just shown, there exists $(p, q) \in A, |p| < |q| \le N$ such that
$$\frac{p'}{q'} = \frac{1}{2\ell + \frac{p}{q}}.$$
The induction step is complete and the result follows. Note that it also follows that the representation of numbers in $A$ as continued fractions like that is unique. Indeed, take a pair $(p', q') \in A$. The pair $(p, q) \in A$ that satisfies (4) and $|p| < |p'|$ is determined uniquely. Uniqueness follows easily.