My question is that if P = NP, then we can solve any NP-hard problems (the one which is NP-complete and the one which is not-NP-complete) by saying that since we have a polynomial time algorithm to solve the NP problems and since the reduction works in polynomial time. Therefore, polynomial to polynomial is polynomial, and so we can solve any NP-hardness problem. For example, we can solve the TSP by calling to an NP-complete problem which have polynomial time since P=NP, and since the reduction is polynomial time, therefore we can solve TSP in polynomial time.
How could disprove this argument? What is the clue to say that NP-hardness(the one which doesn't intersect with NP-complete) can not be solved in polynomial time even if P=NP.
I know that if P = NP, then we can only solve any problem in NP (but not in NP-hard). But I want to know what it makes NP-hard (the one that have no intersection with NP-complete) is not solvable in polynomial time even if P=NP since we have polynomial time reduction and polynomial time algorithm given P=NP.