Such sequences can be described (after discarding a finite segment) via a deterministic recurrence applied to an initial pair $b,c$ of natural numbers; in particular, there are only countably many such sequences.
Indeed, if $\sum \frac{1}{a_n}$ is rational, write $\frac{b_n}{c_n} = \sum_{n' \geq n} \frac{1}{a_n}$, then $\frac{b_n}{c_n} = \frac{1}{a_n} + \frac{b_{n+1}}{c_{n+1}}$. Writing $b_n a_n = c_n + h_n$, we have $b_n | c_n + h_n$, $b_{n+1} = h_n$, and $c_{n+1} = c_n \frac{c_n+h_n}{b_n}$. The condition $\lim_{n \to \infty} \frac{a_n}{a_{n-1}^2}=1$ means that $a_{n+1} = (1+o(1)) a_n^2$, that $\frac{b_n}{c_n} = (1+o(1)) \frac{1}{a_n}$, and thus $\frac{b_{n+1}}{c_{n+1}} = (1+o(1)) \frac{b_n^2}{c_n^2}$, which after some more algebra leads to $h_n = (1+o(1)) b_n$. Because of this, for $n$ sufficiently large one can solve for $\frac{c_n+h_n}{b_n}$ exactly as the nearest integer $\lfloor \frac{c_n}{b_n}+1 \rceil$ to $\frac{c_n}{b_n}+1$, and so the pair $(b_n,c_n)$ iterates via the recurrence
$$ (b_{n+1}, c_{n+1}) = (b_n \lfloor \frac{c_n}{b_n}+1 \rceil - c_n, c_n \lfloor \frac{c_n}{b_n}+1 \rceil).$$
This iteration yields a sequence of the desired form if and only if $\mathrm{dist}(c_n/b_n, {\mathbf Z}) \to 0$ as $n \to \infty$ (or equivalently $b_{n+1} = (1+o(1)) b_n$). If the $b_n$ stayed bounded, then this would force the $b_n$ to eventually become constant, at which point the $c_n$ must be multiples of $b_n$ and one can check that one becomes a Sylvester sequence at that point. So the question comes down to whether there is a solution to the above recurrence in which the $b_{n+1}/b_n$ tend to $1$ but never actually equal $1$ (if $b_{n+1}=b_n$ one can check that $b_m = b_n$ for all $m > n$).
Heuristically, any such sequence should have a zero probability of $b_{n+1}/b_n$ converging to 1 (it requires $c_n/b_n$ to be inexplicably close to an integer for all sufficiently large $n$), which indicates that the answer to this question is "no" [EDIT: I meant "yes"]; but the recurrence looks quite difficult to analyze (especially as the $c_n$ grow doubly exponentially fast) and I see no plausible way to rigorously prove the claim.
