The sets of trees being summed over are not actually different, that's the thing. Let me try to explain it without introducing a notational mess, and hopefully the idea will be clear.
Nonsymmetric cyclic operads are built out of trees equipped with an isotopy class of embedding into the plane. In the free nonsymmetric cyclic operad (equivalently, in the bar construction on a cyclic operad), one forms a sum over all such trees that map down to a given one-vertex tree (a.k.a. corolla) when contracting all internal edges.
Usual nonsymmetric operads are built out of rooted trees equipped with an isotopy class of embedding into the plane, i.e. one of the legs of the trees is marked as "outgoing". In the free nonsymmetric operad, one forms a sum over all such trees that map down to a given rooted corolla when contracting all internal edges.
Now observe that the operation of contracting all internal edges induces a bijection between the legs of the tree and the legs of the resulting corolla. So choosing one of the legs as "outgoing" on the tree is equivalent to making such a choice on the corolla. It follows that the sets of trees being summed over are actually in canonical bijection with each other; the bijection is given by "forgetting which leg is outgoing". In particular the chain complexes in the bar constructions are canonically isomorphic. The "forgetting which leg is outgoing" operation on trees induces the forgetful functor from nonsymmetric cyclic operads to nonsymmetric operads: thus if $P$ is a nonsymmetric cyclic operad, $F$ denotes the forgetful functor from nonsymmetric cyclic operads to nonsymmetric operads, and $B^{cyc}$ resp. $B^{ns}$ denote the two bar constructions, then $F(B^{cyc}(P))$ and $B^{ns}(F(P))$ are canonically isomorphic.
Everything above is true mutatis mutandis for cyclic symmetric operads vs usual symmetric operads.
