Claim that the answer is all $a, b\in\mathbb R$ such that $\log_ab\in\mathbb Q$. First of all, let's prove that if $\log_ab\in\mathbb Q$, then such $P, Q$ exist. Suppose that $\log_ab=\frac dc$ for positive integers $c, d$ (note that since $a, b>1$, there is $\frac dc>0$). $\Rightarrow a^d=b^c$. Consider $P(r)=r^c, Q(r)=r^d$. $P(r)\in A\iff r^c=a^k$ for some $k\iff r^d=b^k$ for some $k\iff Q(r)\in B$. Next, let's prove that if $P(r)\in A\iff Q(r)\in B$, then $\frac{\deg(Q)}{\deg(P)}=\log_ab$. Since the leading coefficients of $P, Q$ are positive, and they are not constants, $P, Q$ must be strictly increasing for all $r>R$ for some $R$ that is large enough. Since $P, Q$ are continuous, unbounded, and strictly increasing in $(R, \infty)$, there exists $R<r_1<r_2<\cdots$ s.t. $P(r_1)=a^k$ for some $k$, and $P(r_{i+1})=P(r_i)a$ for all $i$, and WLOG suppose that for all $r\in(R, r_1)$, $P(r)\notin A$. $\Rightarrow Q(r_1)=b^l$ for some $l$, and $Q(r_{i+1})=Q(r_i)b$ for all $i$. Let $d_p, d_q$ be the degree of $P, Q$, and $c_p, c_q$ be the leading coefficients of $P, Q$, respectively. We know that $\lim_{r\to\infty}\frac{P(r)}{c_pr^{d_p}}=1, \lim_{r\to\infty}\frac{c_qr^{d_q}}{Q(r)}=1$. $\Rightarrow0=\lim_{i\to\infty}\log(P(r_i))-\log(c_pr_i^{d_p})=\lim_{i\to\infty}(k+i-1)\log(a)-d_p\log(c_pr_i)$, and $0=\lim_{i\to\infty}\log(Q(r_i))-\log(c_qr_i^{d_q})=\lim_{i\to\infty}(l+i-1)\log(b)-d_q\log(c_qr_i)$. $\overset{/i}\Rightarrow\lim_{i\to\infty}\log(a)-d_p\frac{\log(r_i)}i=\lim_{i\to\infty}\log(b)-d_q\frac{\log(r_i)}i=0$. $\Rightarrow\frac{\log(a)}{d_p}=\frac{\log(b)}{d_q}=\lim_{i\to\infty}\frac{\log(r_i)}i$. $\therefore\frac{d_q}{d_p}=\log_ab$. Since $d_q, d_p$ are positive integers, $\log_ab$ should be rational. This finishes the proof.

A bit straight forward for P3. Answer: all $a,b>1$ for which $\dfrac{\ln a}{\ln b} \in \mathbb{Q}$. Claim : If for $a,b>1$ such polynomials $P,Q$ exists then $a^{\deg(P)} = b^{\deg(Q)}$.

With the above claim we get that $\dfrac{\ln a}{\ln b} = \dfrac{\deg(P)}{\deg(Q)}\in \mathbb{Q}$ which is the answer of the problem. For $a,b$ with $n\ln a = m\ln b$ let $P(x)=x^{2m}$ and $Q(x)=x^{2n}$. ($2m,2n$ are necessary instead of $m,n$ due to possible negative numbers for $r$).

Proposed by Pouya Mahmoodkhan Shirazi.

Cute! We claim that all $a,b$ such that $\ln a / \ln b$ is rational. Claim 1: $\ln(a)/\ln(b)$ is rational Proof: Let $P = a_nx^n \cdots + a_0$ and $Q = b_kx^k+ \cdots +b_0$, assume after a number $M$; $P(x),Q(x)$ is strictly increasing; now we define a sequence $a_i$, such that for all $i$ $r_i> M$ $P(r_1)=a^t$ for some $t \in N$ $P(r_k)=aP(r_{k-1})$, or $P(r_k)=a^{t+k-1}$ As there is no power of $a$ in $(P(r_k),P(r_{k+1}))$, by IVT we conclude $Q(r_k)=b^{v+k-1}$, now if $a^{n} \neq b^k$, then we have $P(r_k)^{k} > c \cdot Q(r_k)^{n}$ for any constant $c$; which is a contradiction as $c$ is arbitary and degree of both sides is same; now construction is trivial.

All pairs $(a,b)=(c^x,c^y)$ for any real number $1<c$ and positive integers $x$ and $y$ work. To see that these pairs work, take $P(r)=r^{2x}$ and $P(r)=r^{2y}$ (would of missed this if not for mojyla222). For the rest of the solution only consider $r$ positive. We will know show that $\frac{\ln(a)}{\ln(b)}\in \mathbb{Q}$, which is sufficient. Notice that the following must have an infinite number of roots for arbitrarily large $r$ $$\log_a(P(r))-\log_b(Q(r))=\ln\left(\frac{P(r)^{\frac{1}{\ln(a)}}}{Q(r)^{\frac{1}{\ln(b)}}}\right)$$This implies that $P(r)^{\frac{1}{\ln(a)}}=Q(r)^{\frac{1}{\ln(b)}}$ has an infinite number of solutions for arbitrarily large $r$. Let $P$ and $Q$ have degrees of $p$ and $q$. Then $\mathcal{O}(P(r)^{\frac{1}{\ln(a)}})=x^{\frac{p}{\ln(a)}}$ and $\mathcal{O}(Q(r)^{\frac{1}{\ln(b)}})=x^{\frac{q}{\ln(b)}}$. Thus we must have $$\frac{p}{\ln(a)}=\frac{q}{\ln(b)}\Rightarrow \frac{\ln(a)}{\ln(b)}=\frac{p}{q}$$

Really nice problem!, i really liked this one and finally i can use asymptotic tech. I claim that all $a,b$ such that $\text{log}_a(b) \in \mathbb Q$ satisfy the problem. Proof of Necessity: First note that since both $P,Q$ have positive leading coefficients, there exists some $N$ such that for all $x>N$, both $P,Q$ are strictly increasing, now let $c_1>N$ be the minimal such that $P(c_1)=a^k$, then we build the sequence of $c_j$'s in the following way: We let $c_i$ to be the minimal for which $P(c_i)=a^{k+i-1}$ (which all exist as polynomials are continuos), so now due to increasingness we have that $c_j$ for all $j \ge 1$ is an strictly increasing sequence, and due to the problem condition since we have $Q(c_1)=b^{\ell}$ we get that $Q(c_i)=b^{\ell+i-1}$ because by IVT if you ever skipped an exponent some $c_{\kappa}$ would no longer be minimal which would contradict their definition. In a similar way for all large enough $x$ we have that $\frac{P(x)}{x^{\text{deg} P}}$ is strictly decreasing thus $\frac{x^{\text{deg}P}}{P(x)}$ is slowly increasing but it converges to $1$ over the leading coefficient of $P$ (call this $L$) so for some large $x$ this is bounded below by $l>0$ and bounded above by $L>0$, now observe that: $$\lim_{n \to \infty} \sqrt[n]{x_n^{\text{deg}P}}=\lim_{n \to \infty} \sqrt[n]{a^{k+n-1}} \cdot \left(\frac{x_n^{\text{deg}P}}{P(x_n)} \right)^{\frac{1}{n}}=a \implies \lim_{n \to \infty} \sqrt[n]{x_n}=a^{\frac{1}{\text{deg}P}}$$The last step follows because $1=\lim_{n \to \infty} \sqrt[n]{l} \le \lim_{n \to \infty} \left(\frac{x_n^{\text{deg}P}}{P(x_n)} \right)^{\frac{1}{n}} \le \lim_{n \to \infty} \sqrt[n]{L}=1$ so the finish was just Sandwich Theorem. But note that analogously we can get that $\lim_{n \to \infty} \sqrt[n]{x_n}=b^{\frac{1}{\text{deg}Q}}$ therefore $a^{\text{deg}Q}=b^{\text{deg}P}$ which means that $\text{log}_a(b) \in \mathbb Q$ as desired. Proof of Sufficiency: If we have $a^n=b^m$ for integers $m,n$ with $\gcd(m,n)=1$, first note that since $a,b>1$ we get that both $m,n>1$, so now consider $P(x)=x^{2m}$ and $Q(x)=x^{2n}$, they clearly satisfy the problem condition thus we are done .