How is memory managed during C++ copy elision?

#include <iostream>

using namespace std;

class A
{
public :
    A()
    {
        cout<<"constructor is called"<<endl;
    }
    
    ~A()
    {
        cout<<"destructor is called"<<endl;
    }
    
    A(const A &s)
    {
        cout<<"copy constructor is called"<<endl;
    }
};

A beta()
{
    A a;
    cout<<"mem location a : "<<&a<<endl;
    return a;
}

int main(int argc, char** argv) {
    A b = beta();
    cout<<"mem location b : "<<&b<<endl;
    return 0;
}

The above program generates the following output :

constructor is called
mem location a : 0x7ffc12bdaf77
mem location b : 0x7ffc12bdaf77
destructor is called

As far as I understand just a single instance of A was created rather than 2 instances for A a and A b, because of copy elision or return value optimisation. But looking at above program from a memory perspective, object a is inside the stack activation record or stack space of function beta and has a memory location 0x7ffc12bdaf77 . When it returns from beta, copy elision makes object b just same as object a rather than copying it. So, now b also has the address 0x7ffc12bdaf77 . I am unable to understand if b is still a local of the function main and present inside its stack space, how can it occupy a memory outside main's stack space ?