Let $S$ denote all sentences over the signature of $PA$ and $Th(\mathcal{N}) \subseteq S$ denote the theory of true arithmetic. Next, pick your favorite computable bijection $f : S \to \mathbb{N}$. Can there be a set $A \subseteq f(Th(\mathcal{N}))$ that is both recursively enumerable and has nonzero lower asymptotic density? Additionally, does the answer depend on $f$?
I highly suspect there can be no such $A$ for every $f$, but when choosing an appropriate $f$, one might be able to pick such a set consisting only of tautologies/otherwise "trivially true" statements.
This depends wildly on $f$. For example:
Fix some coinfinite recursive $X\subseteq\mathbb{N}$ with asymptotic density $1$ (say, the non-powers of $2$). Now letting $x_i$ denote the $i$th element of $X$ in increasing order, take some recursive $f$ such that $f^{-1}(x_i)=$ "$\underline{i}=\underline{i}$" (where $\underline{n}$ is the numeral corresponding to $n$). This gets a recursive set of true sentences with asymptotic density $1$ under $f$.
Alternately, choosing a recursive $f$ so that $f^{-1}(x_i)=$ "$\neg(\underline{i}=\underline{i})$" produces the opposite effect: the set of all true sentences has asymptotic density $0$ under $f$.
The point is that we can always rearrange a bi-infinite set to get whatever density phenomenon we want, and while $Th(\mathcal{N})$ is extremely complicated there are very simple infinite pieces of it and its complement.