As soon as we have 0, 1 , 1/2 (assumed to satisfy the usual properties) and distributivity, we cannot have an element of additive order 2. Since if
- has order 2, then
$ * = 1 \otimes * = (1/2 + 1/2) \otimes * = 1/2 \otimes * + 1/2 \otimes * = 1/2 \otimes (* + *) = 1/2 \otimes 0 = 0$.
[EDIT: The next sentence goes the other way wrt the question: in the question $\triangleq$ is finer than $\doteq$; here on the contrary $\triangleq$ is assumed to be coarser than $\doteq$] Hence it seems that $\triangleq$ collapses all first-winner games (which have order 2). So I guess that when we take the quotient modulo $\triangleq$ we get a Ring isomorphic to the Surreals.
Hence we should renounce to something. I guess everyone wants to keep 0 and 1, at least. Hence we can
(A) Do as Conway, restrict ourselves to numbers.
(B) Renounce to 1/2. I do not know whether this could work and how could be done exactly. We have to consider some class of games in which there are no "fractions". This should be given a precise meaning, surely we cannot have "divisors of 1", but probably many other games should be taken out.
(C) As above, but discarding only "dyadic" fractions. In this case some identifications are necessary, for example, arguing as above, $1/3 \cdot * $ should be identified with $*$. Again, I do not know whether this could work
(D) Renounce to distributivity. This does not change a lot with the present situation, but leaves open the problem of giving a good definition of the product which does not depend on the representatives.
(E) Consider instead a coarser [EDIT: I meant finer] equivalence relation (classes are smaller)
than Conway's. Then we can find a way to make * have infinite order.
Alas, in this case it is probably hard to mantain the group structure, that is, to have inverses.