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 <del>coarser</del> [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.

EDIT Not exactly, it seems that we can actually have inverses even
with an equivalence relation finer than $\doteq$.
Let $\mathbf{PG}$ be the class of  all
(extensional equivalence classes of)
partizan combinatorial games. Set-theoretical complications can be 
dealt with in the usual way.

$\mathbf{PG}$ is a commutative Monoid wrt $\oplus$ and $0$.
Moreover, the operation $-$ satisfies $-(G \oplus H) =( -G )\oplus (-H)$. 

It is then obvious that if we define $\sim$ by  
$G \sim H$ if and only if there are games $X$ and $Y$ such that
$G \oplus X \ominus X \cong  H \oplus Y \ominus Y $, then
$\sim$ is an equivalence relation on    $\mathbf{PG}$  and
$(\mathbf{PG}/ {\sim}, \oplus/{\sim})$ becomes a Group
with $(-G)/{\sim}$ the additive inverse of $G/{\sim}$. 

I do not know whether  $ \sim$ and $\triangleq$  are the same relation.
However, it is  interesting to notice that
one arrives at $ \sim$ just by group theoretical considerations,
with no need of considering products. 
The relation $\sim$ is the finest relation which makes the quotient
 of $\mathbf{PG}$ a Group, at least if $-$ is supposed to be the inverse.
Were actually 
$ \sim$ equal to $\triangleq$, this would give additional arguments to the 
supposition  that  $\triangleq$
is an interesting relation.

Now for the product.
We need not have 
$(G\oplus H) \otimes K \cong (G\otimes K) \oplus (H\otimes K)$ in
$\mathbf{PG}$, 
for example
as a consequence of the counterexample presented in the question.
However, the classical (and standard)
proof that
$(G\oplus H) \otimes K \doteq (G\otimes K) \oplus (H\otimes K)$
actually shows that
$(G\oplus H) \otimes K \sim (G\otimes K) \oplus (H\otimes K)$,
since in the proof two opposite games annihilate, hence we do not
actually need the full strength of  $\doteq$. Given distributivity, there is no problem in proving associativity (modulo $\sim$).
 
By the definition of $\sim$ and the above formula,
 we easily get that
$ G \sim L$ implies   
$ G  \otimes K \sim L \otimes K$,
for every game $K$. Just use
$(-X) \otimes K = - (X \otimes K )$. 
Thus $\otimes$ passes to the quotient with respect to 
$\sim$.
Hence 
  $(\mathbf{PG}/ {\sim}, \oplus/{\sim},  \otimes/{\sim})$
is a Ring.
This implies that 
 $ \sim$ is finer than $\triangleq$.

Arguing in the same way, we get that
$(\mathbf{PG}/ {\triangleq}, \oplus/{\triangleq},  \otimes/{\triangleq})$
is a Ring. Otherwise, just check that
the $\triangleq$-class of $0$
is an ideal of   $\mathbf{PG}/ {\triangleq}$.
Strictly formally, we should work with 
the $(\triangleq/{\sim})$-class of $0 /{\sim}$.

To check whether  $ \sim$ is the same as $\triangleq$
means to check if the following is true. 
Given any game $G$, if  
$G \otimes H \sim 0$, for every game $H$,
then  $G \sim 0$.

All the above structure definitely deserves further study, so your question
is a really great question!