Quick A: Yes, but with unspecified behavior in case of exception thrown.

Recently on SO:

Using an object without copy and without a noexcept move constructor in a vector. What actually breaks and how can I confirm it?

A vector reallocation attempts to offer an exception guarantee, i.e. an attempt to preserve the original state if an exception is thrown during the reallocation operation. There are three scenarios:

1. The element type is nothrow_move_constructible: Reallocation can move elements which won't cause an exception. This is the efficient case.
2. The element type is CopyInsertable: if the type fails to be nothrow_move_constructible, this is sufficient to provide the strong guarantee, though copies are made during reallocation. This was the old C++03 default behaviour and is the less efficient fall-back.
3. The element type is neither CopyInsertable nor nothrow_move_constructible. As long as it is still move-constructible, like in your example, vector reallocation is possible, but does not provide any exception guarantees (e.g. you might lose elements if a move construction throws).

The normative wording that says this is spread out across the various reallocating functions. For example, [vector.modifiers]/push_back says:

If an exception is thrown while inserting a single element at the end and T is CopyInsertable or is_nothrow_move_constructible_v<T> is true, there are no effects. Otherwise, if an exception is thrown by the move constructor of a non-CopyInsertable T, the effects are unspecified.

I don't know what the authors of the posts you cite had in mind, though I can imagine that they are implicitly assuming that you want the strong exception guarantee, and so they'd like to steer you into cases (1) or (2).