This is false. Moved-from is a "valid but unspecified" state in C++, so perfectly safe, but each type must decide what to do. At the minimum, the destructor must be able to run, because it will be invoked, meaning that the obvious choice is also the only sensible one: letting moved-from be equivalent to the "empty" state.
An empty std::vector does not require that any buffer is allocated. It just has a null data pointer.
It's not true that the only sensible choice for a moved-from object be equivalent to the defaulted constructed one.
If your move constructor doesn't exist then the copy constructor gets called under the language rules, so the sensible default is actually a copy.
Everything else is an optimisation that has a trade-off
A conforming implementation of std::list, for example, can have a default constructor and a move constructor that both allocate a sentinel node on the heap, which is why none of the constructors are noexcept.
If you don't allocate a sentinel on the heap, then moving std::list can invalidate iterators (which is what GNU stdlibc++'s implementation chooses).
I did not say “default-constructed”, because that’s a whole other can of worms.
But yes, the implication of C++ move semantics is that every movable object must also define an “empty” (moved-from) state, so you cannot have something like a never-null unique ptr.
Specifically, it is not allowed for the moved-from object to be inconsistent or to say “using it in any way is UB”, because its destructor will run.
An empty std::vector does not require that any buffer is allocated. It just has a null data pointer.