Would a polynomial-time algorithm for an NP-hard problem implies that P=NP? An NP-hard problem is not in NP. (If it was in NP, it would be an NP-complete problem not NP-hard.)

So my question is: if someone can find a polynomial-time algorithm for an NP-hard problem, would that means that P=NP?

I think yes (I am almost sure) but I can't find the reason why?
Your conclusion is correct: Since an NP-complete problem exists that can be reduced to your NP-hard problem, if you should be able to solve your NP-hard problem in polynomial time the NP-complete problem would be in P, hence P=NP.