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$.

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 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.