Let $A$ be a set of positive integers $\{a_n: n\ge 1\}$ (in increasing order) with positive upper density, that is, $$ \limsup_{n\to \infty}\frac{1}{n}|A \cap [1,n]|>0. $$ Let also $B=\{b_n: n\ge 1\}$ a set of positive upper density.
Is it true that $\{a_{b_n}: n\ge 1\}$ has positive upper density?
Counterexample based on my answer to this question.
The set $$A=(1!,2!]\cup(4!,\ 5!]\cup(7!,\ 8!]\cup(10!,\ 11!]\cup(13!,\ 14!]\cup\cdots$$ has upper density $1$, since $$\frac{|A\cap[1,\ (3n+2)!]|}{(3n+2)!}\ge\frac{(3n+2)!-(3n+1)!}{(3n+2)!}=\frac{3n+1}{3n+2}\to1.$$
The subset $S\subset A$ defined by $$S=(4!,\ 4!+3!]\cup(7!,\ 7!+6!]\cup(10!,\ 10!+9!]\cup(13!,\ 13!+12!]\cup\cdots$$ has density $0,$ since $$\frac{|S\cap[1,\ (3n+1)!+(3n)!]|}{(3n+1)!+(3n)!}=\frac{3!+6!+\cdots+(3n)!}{(3n+1)!+(3n)!}\lt\frac2{3n+1}\to0.$$
However, $S$ has upper density $1$ in $A,$ since $$\frac{|S\cap[1,\ (3n+1)!+(3n)!|}{|A\cap[1,\ (3n+1)!+(3n)!|}\ge\frac{(3n)!}{(3n-1)!+(3n)!}=\frac{3n}{3n+1}\to1.$$
I hope this is good enough. If you insist on putting it in your awful $a_n,b_n,a_{b_n}$ notation, I'm afraid it's going to get ugly.