Yes, the net change in momentum is zero. Which means that, viewed from infinity (i.e., from very far away), the system in question is a single system with an unchanging momentum. The fact that the system starts out, viewed internally, as an object and a star, that then fall together and become a slightly larger star, is irrelevant to the total momentum (and energy) as viewed from infinity.
The case of a black hole is somewhat different, because a black hole is not an ordinary object like a star. As I said before, the hole is made purely of spacetime curvature, and spacetime curvature doesn't "recoil" the way an ordinary object does. You can still assign a momentum to the hole as viewed from infinity, and still look at things from infinity as you could in the case of the object falling into a star, and compute a total momentum for the system that doesn't change. But viewed internally, this will not look like the hole "recoiling" from the impact of the object, the way a star would. It will look like a configuration of spacetime curvature that has a sort of "bend" in it (it's hard to express this in ordinary language).
But in the case of the wormhole, the energy and momentum come out the other end. So, providing nothing hits the "walls" of the wormhole while passing through, it seems to me that the wormhole's energy and momentum (defined as seen from infinity using the asymptotically flat assumption you describe) would be unchanged, at least between the start and end of the process. Possibly during the process, i.e., while the object was inside the wormhole, the wormhole's energy and momentum might appear larger. I haven't tried to calculate it. I'll look up the Visser reference you give when I get a chance.