Consider the two-sorted structure consisting of the natural numbers with the usual ordering together with the set of finite sets of naturals equipped with "$\in$."
I recall seeing a proof that this structure is decidable (that is, its full elementary diagram is computable) - indeed, that the same is true if we include addition - but I can't find it now. I have a sketch of an argument, but it's tedious, and I'd rather have a citation than fill in the details (especially if I'mmisremembering!).
It's worth noting that there are some things we can define in this structure. For example, the set of even numbers can be defined as follows: $n$ is even iff there are finite sets $A,B$ such that
$A\cup B$ is closed downwards,
$0,n\in B$,
for all $b\in B$ we have $b+1\in A$, and
for all $a\in A$, if $a<n$ then $a+1\in B$.
(Similarly, each residue class is definable.) So it's not entirely boring, model-theoretically.