What numbers are in the set $S$?
I don't know how to answer this, but let me elaborate on my comments about your proposed proof strategy. The punchline is that it is very likely that $S$ does contain all positive integers via your strategy, but it is an open problem to prove this.
As you say, a sufficient condition for $S$ to contain a positive integer $N$ is that $N$ can be written in the form $$N=\bigg\lfloor \frac{2^{2^a}}{3^b}\bigg\rfloor.$$ More generally, if you modify your definition of $S$ to start with $x\in S$ and modify your functions to $\alpha(n)=n^y$ and $\beta(n)=\lfloor n/z\rfloor$ for some positive integers $x,y,$ and $z$, then a sufficient condition for $S$ to contain $N$ is that $N$ can be written in the form $$N=\bigg\lfloor \frac{x^{y^a}}{z^b}\bigg\rfloor.$$
I now claim that this is possible for all positive integers $N$ iff every finite sequence of digits appears in the base $y$ expansion of $\log_z x$. So, in your case, that would be the binary expansion of $\log_3 2$. The proof is basically "take $\log_z$ of everything"; I've written up the details below.
Now, the bad news is that questions like this are typically very hard to answer: if you pick some specific nicely defined irrational number, it is usually very difficult to say much about its base expansions. In particular, I'm pretty sure that for all positive integers $x,y,$ and $z$ for which $\log_z x$ is irrational, it is an open problem whether the base $y$ expansion of $\log_z x$ contains all finite sequences.
On the other hand, if you pick a random number (uniformly from some interval), then with probability $1$ it will contain every finite sequence in every base. So it is considered extremely likely that $\log_z x$ contains every finite sequence in base $y$; it would be an amazing discovery if it didn't. In particular, this means it is extremely likely that your set $S$ contains all positive integers.
Of course, since this condition is only sufficient and not necessary for $S$ to contain all positive integers, it is possible that there is an easier proof which does not require solving this open problem. I haven't been able to find any promising approach that doesn't end up reducing to a question of this sort, though.
OK, now here's the proof of the claimed equivalence. Let's write $C=\log_z x$.
First, suppose every finite sequence of digits appears in the base $y$ expansion of $C$, and let $N$ be any positive integer. Choose integers $m$ and $k$ such that $$\log_z N<\frac{m}{y^k}<\frac{m+1}{y^k}<\log_z(N+1).$$ Let the string $s$ be the base $y$ expansion of $m$, with enough initial zeroes added (if necessary) so that $s$ has length at least $k$. This string $s$ appears infinitely many times in the base $y$ expansion of $C$ (since every finite string containing it appears somewhere), and in particular it appears somewhere starting after the decimal point. This means that there exist positive integers $a$ and $b$ such that the base $y$ expansion of $y^aC-b$ is identical to that of $\frac{m}{y^k}$ up to the $k$th digit after the decimal point. That is, $$\frac{m}{y^k}\leq y^aC-b<\frac{m+1}{y^k}.$$ By our choice of $m$ and $k$, this implies that $\lfloor z^{y^aC-b}\rfloor=N$. But $$z^{y^aC-b}=\frac{z^{y^a\log_z x}}{z^b}=\frac{x^{y^a}}{z^b},$$ so this is exactly what we wanted.
Conversely, suppose every $N$ can be written in this form, and let $s$ be a finite string of base $y$ digits. Note that there exists a positive integer $N$ such that the base $y$ expansions of $\log_z N$ and $\log_z(N+1)$ after the decimal point both start with $s$. (This is true because the difference between $\log_z N$ and $\log_z(N+1)$ goes to $0$ as $N$ gets large.) By hypothesis, there exist positive integers $a$ and $b$ such that $$N=\bigg\lfloor \frac{x^{y^a}}{z^b}\bigg\rfloor.$$ This means that $$\log_z \frac{x^{y^a}}{z^b}=y^aC-b$$ is between $\log_z N$ and $\log_z(N+1)$, and thus its base $y$ expansion starts with $s$ after the decimal point. This implies that the string $s$ appears in the base $y$ expansion of $C$ (starting $a$ digits after the decimal point), as desired.