Suppose $V$ is a model of ZF. Within $V$ we have $L$ which is a model of ZFC, furthermore $L[A]$ is a model of choice for every $A\in V$.
Suppose $A=\emptyset$ then clearly $L[A]=L$, furthermore if $A\in L$ then $A\cap L\in L$, therefore $L[A]=L$. Recall also that if $A' = L[A]\cap A$ then $L[A] = L[A']$.
On the other extreme, suppose $A$ is an amorphous set (an infinite set that every subset of it is either finite or co-finite). Consider $A'=A\cap L[A]$, we have that $A'\in L[A]$ which is a model of choice, so $A'$ cannot be infinite - since infinite subsets of amorphous sets are themselves amorphous. Therefore $A'$ is finite, despite not being able to prove that (at the moment anyway) I have a strong intuition that $A'=A\cap L$ and therefore $L[A]=L$.
(this conjecture stems from noticing that amorphous sets are, as the name suggests, amorphous. There is no actual reason that any element would be "preferred" into $L[A]$ over another, unless it was already in $L$. Since there can only be finitely many constructible elements in an amorphous set this somewhat supports my intuition)
Is this conjecture about amorphous sets true? Can it be extended to weaker infinite Dedekind-finite sets? Suppose $L[A]=L$, as Joel points out in his answer this implies $A\cap L\in L$.
- Suppose $L[A]=L$ for every $A\in V$, is there something to say about $V$ and $L$? (in the sense that $V$ is somewhat minimal over $L$ (that is if it has non-well orderable sets, then this is the only difference of $V$ from $L$))
- Suppose $V$ is somewhat larger than the above description (for example, $V$ is the Feferman-Levy model in which $\omega_1$ is singular and the reals have cardinality $\aleph_1$), is there anything to say about sets for which $L[A]=L$? Can we in some sense generate a model $L\subseteq M\subseteq V$ which behaves as described above (some minimality property)?