What does it mean to "solve an equation (or set of equations) for x"? It means, that, assuming the equation(s) hold, you've found out what x must be.
What you've proved is that, **IF** 1 & 2 hold for some x, THEN it must be that x = b'a.
If x = b'a doesn't actually make 1 & 2 hold in your group, then it means that the equations have no solution. It doesn't mean that you made a mistake. You didn't.
Ex: Suppose this was a set of equations in a finite group where 3 does not divide the order of the group, and that a is not equal to b. Could this set of equations have a solution?
No. Since Equation 2 requires that ord(x) divides 3, thus ord(x) is either 3 or 1. But it can't be 3, since that doesn't divide the group order (Lagrange's Thm with <x>).
Thus ord(x) = 1, so x = e. But then ax = b says a(e) = b, so a = b. But that's a contradiction, since a not equal to b by premise.
Thus these equations have no solution in a finite group where 3 does not divide the order of the group, and where a is not equal to b.
Some equations simply don't have solutions - that's true even with the "ordinary" real numbers.