I think I've figured this out, but the answer was not what I expected, so I thought I'd run it by the collective mind to see if I'm missing something.
Feferman proved [1] that there is a subset $A\subseteq\mathbb{N}$ that is hyperarithmetic but where $\{A\}$ is not an arithmetical class. (Actually Feferman proves something considerably stronger.) Rogers [2] outlines the proof in an exercise. Rogers says that a set $A$ is implicitly definable if $\{A\}$ is arithmetical. The proof uses forcing techniques. (See also Boolos et.al.[3].)
It is well-known that the set $V$ of true statements of Peano arithmetic is implicitly definable (see [2] or [3]), and hence hyperarithmetic; the proof amounts to writing out the usual inductive definition of satisfaction in PA augmented with a new predicate $\text{True}(x)$. The proof that Feferman's generic set $A$ is hyperarithmetic is an adaptation of this argument. The question is, why can't we adapt the proof for $V$ to give an implicit definition for $A$?
Here is my current understanding. We can adapt the proof for $V$ to show that the forcing relation $p\Vdash\varphi$ is implicitly definable. Here $\varphi$ is a closed formula in PA augmented with a new predicate $A(x)$, and the condition $p$ gives info on membership in $A$ for finitely many numbers. The implicit definition of $p\Vdash\varphi$ is expressed in PA augmented with a new relation symbol $\text{Forces}(p,x)$. Furthermore, the usual inductive definition of the generic set $A$ can be turned into an implicit definition (using the Forces predicate). So ultimately, if we augment PA with two new predicates, $A(x)$ and $\text{Forces}(p,x)$, then there is a closed formula $\Psi$ in this language such that $\mathbb{N}\models\Psi$ when and only when Forces and $A$ mean what we want. In other words, the class $\{\langle\text{Forces},A\rangle\}$ is arithmetical, and the pair $\langle\text{Forces},A\rangle$ is implicitly definable.
With some simple coding, $\langle\text{Forces},A\rangle$ can be encoded in a subset $B\subseteq\mathbb{N}$, and this $B$ is also implicitly definable. (You can easily code Forces as a subset of $\mathbb{N}$, and then you can interleave Forces and $A$ to get $B$.) Despite the fact that obtaining $A$ from $B$ is nearly trivial, $B$ is implicitly definable and $A$ is not.
This seems to show that the notion of implicit definability is "fragile" in a way that hyperarithmeticity is not. As I said, not what I expected.
[1] S. Feferman, Some applications of the notions of forcing and generic sets, Fundamenta Mathematicae, vol. 56 no. 3 (1965), pp. 325–345.
[2] H. Rogers, Theory of Recursive Functions and Effective Computability, 1967, exercise 16-72 p.452, and Theorem XII sec. 15.1 p.344.
[3] G. Boolos, J. Burgess, R. Jeffrey, Computability and Logic 5th ed. 2007, Ch. 23. This chapter treats both the implicit definability of truth in PA, and the kind of forcing used in Feferman's theorem. However, they skip Feferman's result and go right to Addison's theorem, a missed opportunity IMO.