Do you know about field extensions? Your question is merely a disguised form of the following:
Let F be a field and irreducible over F ( ). Then there is a field F' containing F and an element x in F' such that x2 = 2 and every element of F' is uniquely expressible as s + tx for some s and t in F. Of course the non-zero elements of F' form a group under multiplication. This is precisely your group in either 1 or 2.
I really wouldn't want to prove the group axioms from scratch -- associativity would be a mess just from the definition of *. You said you could find the inverse of any element in case F = Q, but you didn't say what it is. Except for notation, this will also be the inverse when F is GF(5). For either case,
For a and b not both 0, since x^2-2 has no solution in F.