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!