$S$ is recursively enumerable $\iff$ There's some formula $S(x)$ in $PA$ such that $PA\vdash S(\bar{n})$ iff $n\in S$.
Is this claim true? Intuitively I would think so. But I only have a somewhat rough reasoning whose corretness I'm not sure about at all. Any suggestions or different perspectives are welcome.
My reasoning is：
The "$\Rightarrow$" direction
Every r.e. set is definable by a $\Sigma_1^0$-formula. Hence if $S$ is r.e., there's some $S(x)=\exists n_1,...,n_k \phi(n_1,...,n_k,x)$, where $\phi$ has only bounded quantifiers, such that $\mathcal{N}\vDash S(\bar{n})$ iff $n\in S$.
Now we only need to show $PA\vdash S(\bar{n})$ iff $\mathcal{N}\vDash S(\bar{n})$. By soundness of $PA$, $PA\vdash S(\bar{n})$ implies $\mathcal{N}\vDash S(\bar{n})$. On the other hand, $\mathcal{N}\vDash S(\bar{n})$ implies $\mathcal{N}\vDash \phi(\bar{a_1},...,\bar{a_k},\bar{n})$ for some finite numbers $a_1$,...,$a_k$. Since $\phi$ has only bounded quantifiers, this means $\phi(\bar{a_1},...,\bar{a_k},\bar{n})$ must be a valid sentence (Does it?). Hence $PA\vdash S(\bar{n})$ by completeness of FOL.
The "$\Leftarrow$" direction
It's sufficient to show that the set $\{\bar{n}|PA\vdash S(\bar{n})\}$ is r.e.. If we can show it's the domain of some computable function $f$, we are done. Let $F(n)=G(S(\bar{n}))$ where $G(X)$ is the godel number of formula $X$. Now "$m$ is the godel number of a proof of a sentence whose godel number is $n$" is decidable. let $P(m,n)$ be such an algorithm. Then $P(m,F(n))$ is decidable for all $m$ and $n$. Define $f$ by the following algorithm:
for $m,n\ge1$; if $P(m,F(n))=1$, then $f(n)=1$; end
Then $f$ is computable and its domain is $S$.
Q.E.D.