The Grothendieck group of a semiring is a ring. So you are just asking what will happen if you take the multiplicative Grothendieck group of a ring.

Well, it depends on whether you include $0$ in the multiplicative structure! If you include $0$, then $a/b=0a/0b=0/0$ so you get the trivial group. Similarly, inverting zero-divisors will collapse part of the ring. If you only take the Grothendieck group of the multiplicative semigroup of non-zero-divisors, you get the unit group of the total quotient ring. In particular, for an integral domain you get the multiplicative group of the field of fractions. This construction is clearly the same as localization by whatever subset you pick.

If the original semiring is a subsemiring of a field, then it does not matter what order you pick. This is because you can do all the arithmetic inside the field, so the additive Grothendieck group is just the subset generated by a semiring under addition, multiplication and subtraction, and the multiplicative Grothendieck group is the subset generated under addition, multiplication, and division. Applying both gets the field generated by that semiring, regardless of order.

ifyour naturals where to not include 0, I think the order in which you do the 'quotient constructions' would make a difference. With multiplication first, you'd get the positive rationals and then the rationals. While with addition first you'd get the intgers and then just 0. – user9072 Aug 31 '12 at 13:37