A different equivalence relation on partizan combinatorial games The following definitions are fairly standard, but reworded in a way that will be more appropriate for my question (so what follows is fairly long, but should be easy to read for the experts and might even serve as an introduction to the subject):


*

*A partizan combinatorial game is a set $G$ (of positions or states), together with an element $x_0 \in G$ (the starting position or initial state) and two relations, $L, R \subseteq G^2$ which we call the blue and red edges of the graph $(G, L\cup R)$ (an edge which is both blue and red can be called green), such that the graph in question is well-founded (=progressively finite, =there is no infinite sequence $z_0,z_1,z_2,\ldots$ of elements of $G$ with $(z_i,z_{i+1}) \in (L\cup R)$ for each $i$).  The game is played as follows: starting with $x = x_0$, each player (called Left and Right) selects an edge $(x,y)$ in his respective set ($L$ for Left and $R$ for Right), replacing the position $x$ by its out-neighbor $y$; the player who cannot play loses (something which always happens in finite time by the well-foundedness assumption).

*Define the extensional collapse of a game $G$ as follows: first, remove all vertices unreachable from the starting position; then, repeatedly (transfinitely) identify two vertices $x,x'$ which have the same sets of blue-neighbors and the same sets of red-neighbors (i.e., identify $x,x'$ if $\{y : (x,y) \in L\} = \{y : (x',y) \in L\}$ and the same is true for $R$).  Call two games $G,G'$ extensionally equivalent when they have isomorphic extensional collapses and write $G \cong G'$.  For example, any two games in which no player has any legal play ($L=R=\varnothing$) are extensionally equivalent to the trivial game $0$ with only one position.  Essentially everything will be studied up to extensional equivalence.

*The sum $G'' := G \oplus G'$ of two games $G,G'$ is the Cartesian product $G\times G'$ with starting position $x''_0 := (x_0,x'_0)$ and edge relations $L'' := (L\times\Delta_{G'}) \cup (\Delta_G\times L')$ and similarly for $R''$, where $\Delta$ stands for the diagonal: i.e., playing a move in $G\oplus G'$ consists of playing a move in $G$ xor playing one in $G'$.  Clearly this is well-founded, and clearly this operation is compatible with extensional equivalence, and it is commutative and associative (up to extensional equivalence, in fact, even isomorphism).  The opposite $-G$ of a game $G$ is the same set $G$ but with $L$ and $R$ exchanged; we write $G \ominus G'$ for $G \oplus (-G')$.

*A game $G$ is Conway-positive when Left has a winning strategy no matter who plays first, Conway-negative when Right has (i.e., $-G$ is positive), and Conway-zero when the second player has a winning strategy.  Two games $G,G'$ are Conway-equal when $G-G'$ is Conway-zero.  Let me write $G \doteq G'$ for this.  It is not hard to see that $\doteq$ is an equivalence relation, which is weaker (=larger) than extensional equivalence.

*The product $G'' := G \otimes G'$ of two games $G,G'$ is defined informally as follows.  Its positions will be finite combinations of "plus" and "minus" tokens on the vertices of $G \times G'$.  The starting position has a single "plus" token on $(x_0,x'_0)$ (the starting positions of $G$ and $G'$).  At each turn, each player chooses a single token ("plus" or "minus"), at $(x,x')$ say, and chooses edges $(x,y)$ from $G$ and $(x',y')$ from $G'$: for a "plus" token, Left must choose two blue edges or two red edges (i.e., $(x,y) \in L$ and $(x',y') \in L'$, or else $(x,y) \in R$ and $(x',y') \in R'$) and Right must choose a blue edge and a red edge; for a "minus" token, it's the reverse (Left chooses a blue edge and a red edge, and Right chooses two blue edges or two red edges); in either case, the token that was at $(x,x')$ is removed and three new tokens are created: two of the same sign as the original token at $(x,y')$ and $(y,x')$, and one of opposite sign at $(y,y')$.  In brief, we move from $x\otimes x'$ to $(x\otimes y') \oplus (y\otimes x') \ominus (y\otimes y')$.
The product has some nice properties: it is commutative and associative (up to extensional equivalence), it is distributive over the sum (ditto), has $1$ (defined below) as unit element (again, up to extensional equivalence) and $-G \cong (-1) \otimes G$.
For impartial games (those with $L=R=:E$), Conway-equality is exactly the equality of the Grundy ordinals (the Grundy ordinal of an impartial game $(G,x)$ being recursively defined as the least ordinal not equal to the Grundy ordinal of $(G,y)$ for any out-neighbor $y$ of $x$, i.e., $(x,y) \in E$), and the sum and product operations are compatible with this equivalence, and define over the ordinals two operations called "nim sum" and "nim product" which have plenty of nice properties (e.g., each infinite cardinal is an algebraically closed field of characteristic $2$; Conway calls these "nimbers", a portmanteau of "nim" and "number").
Sadly, whereas the sum is always compatible with Conway equality (if $G_1 \doteq G_2$ then $G_1 \oplus G' \doteq G_2 \oplus G'$), product is not in general.  For a counterexample, leg $1$ be the game with two positions $0,1$ and a blue edge between $1$ and $0$, let $2$ be the game with three positions $0,1,2$ and blue edges $L = \{(1,0),(2,0),(2,1)\}$, and let $1\oplus 1$ be the same game as $2$ but without the $(2,0)$ edge (this is, indeed, extensionally equivalent to the sum $1\oplus 1$ as defined above): then $1\oplus 1 \doteq 2$ yet $(1\oplus 1)\otimes * \cong * \oplus * \doteq 0$ is not Conway-equal to $2 \otimes * =: * 2$, where here $*$ and $* 2$ are the same games as $1$ and $2$ but with green (=both blue and red) edges instead of just blue (multiplying by $*$ is extensionally the same as making the game impartial by making all edges green): the Grundy ordinals of $*$ and $*2$ are $1$ and $2$ respectively, so they are not Conway-equal.
So there are two obvious ways to try to solve this problem:


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*Restrict the definition of the product $G \otimes N$ for certain games $N$ for which $G \doteq G'$ implies $(G\otimes N) \doteq (G'\otimes N)$.  This is Conway's approach in On Numbers and Games, the games $N$ in question being called "numbers" (I'm not sure whether all the games for which $G \doteq G'$ implies $(G\otimes N) \doteq (G'\otimes N)$ are numbers, but numbers satisfy this property).  This gives a nice theory where the numbers are a field and the games a partially ordered group, and we can multiply games and numbers, but we can't multiply two games.

*But the other would be to change the equivalence relation $\doteq$, and define a stronger (=smaller, =finer) equivalence relation, say $\triangleq$, by $G\triangleq G'$ iff $(G\otimes H) \doteq (G'\otimes H)$ for all games $H$ (i.e. not only does the second player win $G' \ominus G$, but in fact $(G'\ominus G) \otimes H$ for all games $H$; in particular, this is required for $H = *$, which demands that the impartial games $G \otimes *$ and $G' \otimes *$ have the same Grundy ordinals).  It seems that this should still give the games (up to a certain size…) a ring structure, of which the nimbers are a factor or something like that.
My question (at last) is whether this second approach has been studied, and whether it leads to an interesting theory or whether on the contrary there is something obviously wrong with it.
 A: 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.
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! 
