The functions which preserve asymptotic equivalence have turned out to be one of a vast number of generalizations of Karamata's regularly varying functions. Paper "On some extensions of Karamata's theory and their applications" by Buldygin et. al. introduces pseudo-regularly varying (PRV) functions which are, in fact, exactly that measurable functions, not necessarily tending to infinity, which preserve asymptotics at $+\infty$ in my sense.
The paper contains many properties of PRV functions and conditions for a function to be PRV. For example, PRV fuctions are characterized as being eventually ($\forall x>x_0$) of the form
$$
f(x) = \exp \bigg( a(x) + \int\limits_{x_0}^x b(t) \frac{dt}{t} \bigg),
$$
where $a$ and $b$ are bounded measurable functions with $\lim\limits_{c\rightarrow 1}\limsup\limits_{x\rightarrow +\infty} \lvert a(cx)-a(x) \rvert = 0$.
So the reference request is fulfilled, I'll return to finer class $\mathcal A$ described in the question and answer the "mini-questions".
Besides the above, we can charactirize $\mathcal A$ in terms of uniform continuity, although this characterization is not so constructive.
Proposition. A continuous function $f\colon \mathbb R_+ \rightarrow \mathbb R_+$ with $\lim \limits_{x \rightarrow +\infty} f(x) = +\infty$ belongs to $\mathcal A$ if and only if function $F(t) = \log f(e^t)$, correctly defined on some ray $[T,+\infty)$, is uniformly continuous.
Proof. Note that $\log f(x) = F(\log x)$.
1) To prove the "if" part, pick $\{a_n\}_{n=1}^\infty$ and $\{b_n\}_{n=1}^\infty$ as in the definition of $\mathcal A$. Since $\lim\limits_{n\rightarrow\infty} \log\tfrac{a_n}{b_n} = 0$ and $F$ is uniformly continuous,
\begin{align*}
\left| \log \frac{f(a_n)}{f(b_n)} \right| &= \lvert \log f(a_n) - \log f(b_n) \rvert = \lvert F(\log a_n)-F(\log b_n) \rvert = \\ &= \lvert F(\log b_n+\log\tfrac{a_n}{b_n}) - F(\log b_n) \rvert \leqslant \sup\limits_{t \geqslant T} \lvert F(t+\log\tfrac{a_n}{b_n}) - F(t) \rvert \xrightarrow{\;n\rightarrow\infty\;} 0,
\end{align*}
so finally $f(a_n) \sim f(b_n)$ as $n$ tends to infinity.
2) To prove the "only if" part, suppose that, coversely, $F$ is not uniformly coninuous. That means there exists $\varepsilon > 0$ and two sequences $\{t_n\}_{n=1}^\infty, \{s_n\}_{n=1}^\infty \subseteq [T, +\infty)$, such that
$$
\lim\limits_{n\rightarrow\infty} (t_n-s_n) = 0 \quad\text{and}\quad \forall n\in \mathbb N\ \; \lvert F(t_n) - F(s_n) \rvert \geqslant \varepsilon.
$$
It's necessary that $\{t_n\}_{n=1}^\infty$ is unbounded, because by Cantor's theorem $F$ is uniformly continuous on each finite interval. Hence, taking subsequence if needed, we can assume that $t_n$ (and so $s_n$) goes to infinity.
Define $a_n := \exp(t_n), \; b_n := \exp(s_n)$. We have $\lim\limits_{n\rightarrow\infty} \tfrac{a_n}{b_n} = \lim\limits_{n\rightarrow\infty} \exp(t_n-s_n) = 1$ and
$$
\bigg\lvert \! \log \frac{f(a_n)}{f(b_n)} \! \bigg\rvert = \lvert F(\log a_n)-F(\log b_n) \rvert = \lvert F(t_n)-F(s_n) \rvert \geqslant \varepsilon \;\;\; \forall n \in \mathbb N,
$$
what contradicts with $f \in \mathcal A$.
From this characterization it follows that a functions from $\mathcal A$ grows at most like power function. (It's also follows from the exponent expression for PRV functions.)
Corollary. If $f \in \mathcal A$, then $f(x) = O(x^\alpha)$ for some $\alpha > 0$ as $x \rightarrow +\infty$.
Proof. Take $F$ like in above proposition. Since $F$ is uniformly continuous, $F(t) \leqslant \alpha t \;\;\; \forall t \geqslant T$ with some $\alpha,T \geqslant 0$ (see this question). Then
$$
f(x) = \exp\big(F(\log x)\big) \leqslant \exp(\alpha \log x) \leqslant x^\alpha \quad \forall x \geqslant e^T.
$$
And finally, we can indeed replace equivalent sequences in the definition of $\mathcal A$ with such $C^\infty$-smooth functions on $\mathbb R_+$ and get the same class. It's straightforward, we just need to construct smooth function with given values in, say, $\mathbb N$.
We may also restrict the sequences from the definition to be strictly increasing, hence each real sequence not bounded above admits a strictly increasing subsequence. And thus the smooth functions from the alternative definition can be all stricly increasing. It's bit more tricky and this is crucial for l'Hopital's rule application to prove
Proposition. If $f, g \in \mathcal A$, then $F(x)=\int_0^x f(t)dt$ and $G(x) = \int_0^x g(t)dt$ lie in $\mathcal A$ and, moreover, $F \sim G$ at infinity.
You made a nice guess!
Suppose, $k=2^{\sqrt{\log_2{n}}}$ taking $\log_2$ both side, $$\log_2{k}=\sqrt{\log_2{n}}$$
On the other hand take $\log_2$ for $\log_2{n}$ gives $\log_2{\log_2{n}} $. Check the behaviors of $\log_2{x}$ and $\sqrt{x}$,(see here) and as we are focusing on asymptotic behavior of functions, we will get $\log_2{x}<\sqrt{x}$, for large values of $x$(for any $x>16$) . Putting $x=\log_2{n}$ gives $\log_2{\log_2{n}}$ is smaller than $\sqrt{\log_2{n}}$. As, $\log_2$ is an strictly increasing function, hence, $\log_2{\log_2{n}}<\sqrt{\log_2{n}}=\log_2{k}$ gives $\log_2{n}<k$. Hence, $k=2^{\sqrt{\log_2{n}}} $.
Similarly, take $\log_2$ for $n^{\frac{1}{3}}$. As, $\log_2{n^{\frac{1}{3}}}= \frac{1}{3}\log_2{n}>\sqrt{\log_2{n}}$, we will have $n^{\frac{1}{3}}$ after $2^{\sqrt{\log_2{n}}}$ in your list(due to strictly increasing property of $\log_2$). So, the list will look like this:
$\log_{2}n,\quad 2^{\sqrt{\log_2{n}}}, \quad n^{\frac{1}{3}}, \quad n^{5}, \quad 10^{n}, \quad n^{n}$
Best Answer
That $\mathcal O(1.5^n)$ seems to grow slower than $\mathcal O(n^2)$ (or $\mathcal O(n^3)$) is an illusion. Exponential functions with exponents greater than 1 grow faster than all polynomial functions, but only eventually.
This is the key word: growth rates describe limiting behaviour, not behaviour at a particular point. In particular, the point where $1.5^n$ overtakes $n^2$ in growth was off your graph, and you missed it.