The only possibility is $a=b$. Claim. There exist infinitely many $n$ such that both intervals $(na, na+a]$ and $(nb,nb+b]$ contain only composite numbers. Proof: For the sake of contradiction, suppose there exists $N\in\mathbb{N}$, and for all $n\ge N$ there is a prime either in $(na, na+a]$ or in $(nb,nb+b]$ (or in both). Then for any $k\in \mathbb{N}$, in some of the intervals $(Na, Na+ka]$ or $(Nb, Nb+kb]$ there exist at least $k/2$ prime numbers. It means the prime numbers have a positive density, which contradicts the prime number theorem (PNT). $\blacksquare$ For those $n$, satisfying the above claim, we have $\pi((n+1)a)=\pi(na)$ and $\pi((n+1)b)=\pi(nb)$, hence $$\frac{P(n)}{Q(n)}=\frac{P(n+1)}{Q(n+1)}$$for infinitely many naturals $n$. It yields $$P(z)Q(z+1)=P(z+1)Q(z)\,,\,\forall z\in \mathbb{C}\,\,\,\,\,\,\,\,\,(1)$$If $z_0\in\mathbb{C}$ is a root of $P(z)$ with maximal $\mathrm{Re}\,(z)$, the above identity implies $z_0$ is also root of $Q(z)$. Note that $P_1(z):=\frac{P(z)}{z-z_0}$ and $Q_1(z):=\frac{Q(z)}{z-z_0}$ also satisfy $(1)$. Thus, applying that operation several times, we get $P(z)=cQ(z)$ for some constant $c\in \mathbb{C}$. By PNT $\pi(an)\sim \frac{an}{\ln(an)}$, so it easily follows $c=\frac{a}{b}$. Hence $$\frac{\pi(an)}{\pi(bn)}=\frac{a}{b}$$It gives $$\frac{\pi(an+a)-\pi(an)}{\pi(bn+b)-\pi(bn)}=\frac{a}{b}$$Let $n=kb!$. Then $bn+2,bn+3,\dots,bn+b$ are all composite numbers. Consider the sequence $kb\cdot b!+1, k=1,2,\dots$. By the Dirichlet theorem, there exists $k\in\mathbb{N}$ such that $kb\cdot b!+1$ is a prime. For that $k$ we get $\pi(bn+b)-\pi(bn)=1$. It yields $a/b$ is integer. With the same trick, we can find $n$ such that $\pi(an+a)-\pi(an)=1$, which yields $b/a$ is integer. Hence, the only possibility is $a=b$. In this case, the condition of the statement obviously holds. Remark. I like the kind of problems, like this one, where ideas from different branches overlap. At the beginning we have some analytic argument, then a kind a functional(polynomial) equation, and finally a NT argument. EDIT: $(1,1)$ is not a solution, since $\pi(1)=0$. (Thanks, XbenX !)

Is there any other answers(don't use Dirichlet theorem ) can prove if $$\frac{\pi(an)}{\pi(bn)}=\frac{a}{b}$$then a=b?