Yes, it works exactly like this, this is a demo on godbolt [0]. rdi stores the pointer in both cases, makeS1() uses RVO, makeS2() takes it explicitly and constructs with placement new.

I will say before testing this i didn't realize the RVO calling convention was to return the pointer you pass in, but apparently so. If makeS2() returned void, it's just a tail call to the constructor, but makeS1() has to spill rbx and use it to save the pointer.

[0]: https://godbolt.org/z/ovd1n99P8

No, all you're showing in that example is that a pointer is passed as part of the ABI. You're not showing that RVO relates to that in any way whatsoever. If you write the same function in a manner that (N)RVO can't kick in, does the pointer no longer get passed?

The reason this should sound dubious is that you're suggesting the caller needs to know the callee's body in order to know how to call it, but it should be possible for the two to be compiled entirely independently, and in fact mutual recursions should be fine too. After all, the callee knows where the return value has to land either way, regardless of when/how it's constructed or destroyed.