No. It's not possible to give such example. Not explicitly, not in the generality implied by the question, anyway.
The reason is that it is consistent that $\bf V=L$. In such case it is not possible to give a counterexample.
However suppose that $M$ is a countable transitive model of ZFC+$\bf V=L$. We can extend $M$ by forcing and add new sets. Let $N=M[G]$ a generic extension of $M$, then $N$ is also a countable transitive model. Therefore $M=\mathbf L^M=\mathbf L^N$. But $N$ has new sets which are not in $M$, and therefore these sets satisfy the required property.
Now consider working internally in $N$, then $N=\bf V$ now, and $M=\bf L$. So we have some generic set $G\in N\setminus M=\bf V\setminus L$. However describing in explicit details such set would be impossible for the same reason it is impossible to describe a well-order of $\mathbb R$ without using the axiom of choice. It is simply consistent that there is none, unless we assume this is not the case.
As Andres Caicedo comments, there are plenty of sets that are not in $\bf L$ but their existence requires us to assume additional axioms. For example if $\kappa$ is a measurable cardinal and $\cal U$ is a $\kappa$-complete ultrafilter on $\kappa$, then $\cal U\notin\bf L$.
There are also axioms that when assumed assure that there are sets not in $\bf L$ which do not require an additional consistency strength. For example if we assume that $CH$ fails then $2^{\aleph_0}>\aleph_1$, and then we must have that almost all the real numbers are not in $\bf L$.