The definition of number and integers
The definition of "number" in Combinatorial Game Theory is not "a game $G$ such that every right option is greater than every left option". For example, with $*\cong\{0|0\}$ and $\uparrow\cong\{0|*\}$, the game $\{0|\uparrow\}$ satisfies your condition (as $\uparrow>0$), but is not a number.
One definition of "number" could be Definition A: "a game $G$ such that all options are numbers and every right option is greater than every left option", but that's stronger than necessary.
I prefer defining "number" as Definition B: "a game $G$ such that all options are numbers and no right option is less than or equal to any left option" and then proving it satisfies the stronger definition.
Note that the natural numbers games you defined (e.g. $0=\{|\}$,$1=\{0|\}$,...) are numbers in the sense of either definition by an immediate induction: their option is a number by induction and they don't have a right option to violate the inequality clause.
Also note that the negative of a number (by either definition) is automatically a number (by the same definition), so that gives us the rest of the integers. I'll talk about the dyadic fractions (rationals whose denominator is a power of 2) later.
Simplicity Rule
The simplicity rule will take some work to reach. I'll use definition B in what follows and prove that numbers satisfy definition A along the way.
Lemma 1: Numbers are between their options
Proof: We'll prove that a number $G$ is greater than all of its left options; the proof that it's less than all of its right options is exactly analogous. Let $G$ be a number, and assume, as an inductive hypothesis, that lemma 1 is true for all of the options of $G$. Let $G^L$ be a left option of $G$. Then to check $G>G^L$, we need to see that Left can always win $G-G^L$. If Left goes first, she can just move to $G^L-G^L=0$ and win. If Right moves first to some $G^R-G^L$, then since $G$ is a Number, we know that $G^R\nleq G^L$ (by definition B) so that Left wins $G^R-G^L$ going first. Finally, if Right moves first to some $G+\left(-G^L\right)^R=G-\left(G^{LL}\right)$ (by this I mean that a right option of $-G^L$ is the negative of a left option of $G^L$), then Left can win by moving to $G^L-G^{LL}>0$ (inductive hypothesis applied to $G^L$). $\square$
Corollary 1
Now we can see that definition B implies definition A. By lemma 1, which only assumed definition B, we have $G^L<G$ for all left options, and $G<G^R$ for all right options. By transitivity of $<$, we can conclude that for all pairs of options, $G^R>G^L$, as desired. $\square$
Lemma 2: Numbers are closed under sum
Proof: Let $G$ and $H$ be numbers. Recall that $G+H=\left\{\mathcal{G}^L+H,G+\mathcal{H}^L\left|\mathcal{G}^R+H,G+\mathcal{H}^R\right.\right\}$. Assume, for sake of contradiction, that some right option of $G+H$ is less than or equal to a left option. Case I: Suppose $G^R+H\le G^L+H$. Then $G^R\le G^L$ which is impossible since $G$ is a number. Case II: Suppose $G^R+H\le G+H^L$. By lemma 1, $H^L<H$, so $G+H^L<G+H$. Analogously, $G+H<G^R+H$. But then we have $G+H<G^R+H\le G+H^L<G+H$, which is a contradiction. Cases III and IV are analogous. $\square$
Lemma 3: Numbers are pairwise comparable
Proof: If $G$ and $H$ are numbers, then $-H$ is a number by the remark in the first section, and hence $G-H$ is a number by lemma 2. Since $G$ being incomparable with $H$ is equivalent to $G-H$ being a first-player win (an $\mathcal{N}$-position), it suffices to show that no number is a first-player win. Suppose for sake of contradiction that a number $G$ is a win for the first player. Then choose winning moves $G^L$ and $G^R$. To be winning moves, we must have $G^L\ge0$ and $G^R\le0$, so that $G^R\le G^L$, which is a contradiction. $\square$.
Lemma 4: Proto-Simplicity
Definition: A game $H$ "fits" for a game $G$ if $\mathcal{G}^L\ngeq H\ngeq\mathcal{G}^R$; that is, $H$ fits for $G$ if $H$ lies strictly between (or is incomparable to) the left and right options of $G$. With this definition, lemma 1 implies that a number fits for itself.
Lemma Statement: If $H$ is a number, and $H$ fits for $G$ but none of $H$'s options fit for $G$, then $G=H$.
Proof: We need to show that the second player has a winning strategy on $G-H$. If Left makes the first move to $G^L-H$, then since $H$ fits for $G$, $G^L-H\ngeq0$ and Right can win. Now suppose Left makes the first move to $G-\left(H^R\right)$. Since none of $H$'s options fit for $G$, $H^R$ doesn't fit for $G$, and there must be a witnessing (in)equality of the form $G^L\geq H^R$ or $G^R\leq H^R$. By lemma 1, $H^R>H$, and since $H$ fits for $G$, $G^L<H<H^R$, so that $G^L\geq H^R$ would be impossible. Therefore, there must be an option $G^R\leq H^R$. Thus, Right can win by moving from $G-\left(H^R\right)$ to $G^R-\left(H^R\right)\le0$. The two cases in which Right makes the first move are analogous. $\square$
Lemma 5: Simplest fitting numbers
Statement: If $G$ is a game, and any number fits for $G$, then there is a lowest birthday among birthdays of numbers $H$ that fit for $G$.
Proof: This follows immediately from the fact that the ordinals are well-ordered. $\square$
Theorem: Simplicity Rule
Statement: If $G$ is a game, and any number fits for $G$, then $G$ is equal to any minimal-birthday fitting-number.
Proof: If $H$ has the lowest birthday of any fitting number, no option of $H$ could fit, so the result follows from lemma 4. $\square$
Corollary 2
In particular, if $G$ is equal to a number $H$, lemma 1 shows that $H$ fits for itself (and hence $H$ fits for $G=H$), so $G$ is equal to any minimal-birthday fitting-number by the simplicity rule. $\square$
Note that the simplicity rule doesn't say anything about uniqueness: the number $\omega\cong\{0,1,2,3,\ldots|\}=\{0,2,4,\ldots|\}$, but both forms have minimal birthday $\omega$ (the ordinal).
All finite-birthday numbers are equal to a dyadic
Proof: Suppose that $G$ is a number with finite-birthday and all numbers of earlier birthday are equal to a dyadic. Then $G$ has finitely many options, all dyadic. Since the dyadics (and in fact all numbers, by lemma 3) are pairwise comparable, we can assume $G$ is in canonical form with at most one left option and at most one right option, both necessarily dyadic. If an integer fits, then by applying lemma 4 and induction, $G$ is equal to the smallest absolute value integer that fits. If no integer fits, then there's a unique dyadic of smallest denominator that fits, and $G$ is equal to that.
How do we get $\mathbb Q$ and $\mathbb R$
You can get elements of $\mathbb R$ that are not-dyadic on day $\omega$ by using Dedekind cuts with dyadics. In other words, if $r$ is a real that is not dyadic, then $r=\left\{\text{dyadics}<r\left|\text{dyadics}>r\right.\right\}$.
There's also a nice implicit definition of the reals: a number $G$ is real if $G$ is bounded by integers ($-n<G<n$ for some natural $n$) and $G$ is "simply approximated by dyadics" in the sense that $G=\{G-q|G+q\}$ where $q$ ranges over the positive dyadics.
Some of this was based on old blog posts I made as I was working through Winning Ways/On Numbers and Games, but a couple tricky bits are based on Aaron N. Siegel's Combinatorial Game Theory.
Overall, that PDF skips over some key clarifications about notation that may have led to some confusion. I would not recommend learning from it without external support like lectures or another more detailed text.
for any value $x$ we have $x+∗=\{x\mid x\}$.
The context in the PDF is that this holds when $x$ is a number. If $x$ is some other game (like $\{0\mid*\}$), this equation can fail.
Given the previous, it seems logical that both $∗+∗=\{∗\mid∗\}=0$ and
$\emptyset+∗=\{\emptyset\mid\emptyset\}=0$ are true.
$∗+∗=\{∗\mid∗\}=0$ is true. But we need to take more care when evaluating something like $\emptyset+∗$. Since every game is an ordered pair of games, $\emptyset$ is not a game. However, there is a convention, used-but-not-explained in the addition section of that PDF for expressions that look like adding a set to a game. If $S$ is a set of games and $g$ is a game, $S+g$ is a shorthand for "the set of all games of the form $s+g$, for some $s$ in $S$". So $\emptyset+∗$ is a set, not a game like $0$. In particular, it's the set $\emptyset$.
This looks to me like $∗$ and $\emptyset$ could be equivalent because given $x+∗=0$,
$∗$ and $\emptyset$ both seem like valid solutions for $x$.
As discussed above, $\emptyset+*$ is not $0$ -- it's not even a game. But you are right that "$g+*=0$ and $h+*=0$" would imply "$g=h$". In fact, you could add $*$ to both sides of $g+*=0$ to find $g+*+*=*$ so that $g+0=*$ and $g=*$. This idea works in general; negatives of games are unique (up to equality).
However I have not read anywhere that ∗ and ∅ are equivalent, so I am
not sure if they are or if they are not.
Just to emphasize one more time: $*$ is a game and $\emptyset$ is a set. They're different sorts of things.
Best Answer
Notation Issue
In Combinatorial Game Theory, integers like $4$ each denote a particular game or game value, like $\{3\mid\,\}$ (note that we don't write $\varnothing$ in this notation).
As nombre pointed out in the comments, the written equations like "$n=\{\frac{n}{2}\mid\varnothing\}$" are rarely/never true under standard notation for combinatorial games.
If you don't intend to be referencing the standard meanings of $n$, $3n+1$, and $\frac{n}{2}$ in this notation, you should either have a gigantic disclaimer that the usual notation does not apply, or just use something else for the games you would like to describe.
I'll use $g(n)$ where you have $n$, etc. So we have $g(1)=\{\,\mid\,\}$, $g(n)=\{g(n/2)\mid\,\}$ for even $n$, and $g(n)=\{\,\mid g(3n+1)\}$ for odd $n>1$. Technically, this is only a valid definition for all $n$ if the Collatz conjecture is true.
Example Numeric Values
Let's start building a table and see if we see any patterns. $g(1)=\{\,\mid\,\}=0$. $g(2)=\{g(1)\mid\,\}=\{0\mid\,\}=1$. $$\begin{align}g(3)&=\{\,\mid g(10)\}\\&=\{\,\mid \{g(5)\mid\,\}\}\\&=\{\,\mid \{\{\,\mid g(16)\}\mid\,\}\}\\&=\{\,\mid \{\{\,\mid \{g(8)\mid\,\}\}\mid\,\}\}\\&=\{\,\mid \{\{\,\mid \{\{g(4)\mid\,\}\mid\,\}\}\mid\,\}\}\\&=\{\,\mid \{\{\,\mid \{\{\{g(2)\mid\,\}\mid\,\}\mid\,\}\}\mid\,\}\}\\&=\{\,\mid \{\{\,\mid \{\{\{1\mid\,\}\mid\,\}\mid\,\}\}\mid\,\}\}\\&=\{\,\mid \{\{\,\mid \{\{2\mid\,\}\mid\,\}\}\mid\,\}\}\\&=\{\,\mid \{\{\,\mid \{3\mid\,\}\}\mid\,\}\}\\&=\{\,\mid \{\{\,\mid 4\}\mid\,\}\}\\&=\{\,\mid \{0\mid\,\}\}\\&=\{\,\mid 1\}\\&=0\end{align}$$ $g(4)=2$, $g(5)=0$, $g(6)=\{g(3)\mid\,\}=\{0\mid\,\}=1$. $$\begin{align}g(7)&=\{\,\mid g(22)\}\\&=\{\,\mid \{g(11)\mid\,\}\}\\&=\{\,\mid \{\{\,\mid g(34)\}\mid\,\}\}\\&=\{\,\mid \{\{\,\mid \{g(17)\mid\,\}\}\mid\,\}\}\\&=\{\,\mid \{\{\,\mid \{\{\,\mid g(52)\}\mid\,\}\}\mid\,\}\}\\&=\{\,\mid \{\{\,\mid \{\{\,\mid \{g(26)\mid\,\}\}\mid\,\}\}\mid\,\}\}\\&=\{\,\mid \{\{\,\mid \{\{\,\mid \{\{g(13)\mid\,\}\mid\,\}\}\mid\,\}\}\mid\,\}\}\\&=\{\,\mid \{\{\,\mid \{\{\,\mid \{\{\{\,\mid g(40)\}\mid\,\}\mid\,\}\}\mid\,\}\}\mid\,\}\}\\&=\{\,\mid \{\{\,\mid \{\{\,\mid \{\{\{\,\mid \{g(20)\mid\,\}\}\mid\,\}\mid\,\}\}\mid\,\}\}\mid\,\}\}\\&=\{\,\mid \{\{\,\mid \{\{\,\mid \{\{\{\,\mid \{\{g(10)\mid\,\}\mid\,\}\}\mid\,\}\mid\,\}\}\mid\,\}\}\mid\,\}\}\\&=\{\,\mid \{\{\,\mid \{\{\,\mid \{\{\{\,\mid \{\{1\mid\,\}\mid\,\}\}\mid\,\}\mid\,\}\}\mid\,\}\}\mid\,\}\}\\&=\{\,\mid \{\{\,\mid \{\{\,\mid \{\{\{\,\mid \{2\mid\,\}\}\mid\,\}\mid\,\}\}\mid\,\}\}\mid\,\}\}\\&=\{\,\mid \{\{\,\mid \{\{\,\mid \{\{\{\,\mid 3\}\mid\,\}\mid\,\}\}\mid\,\}\}\mid\,\}\}\\&=\{\,\mid \{\{\,\mid \{\{\,\mid \{\{0\mid\,\}\mid\,\}\}\mid\,\}\}\mid\,\}\}\\&=\{\,\mid \{\{\,\mid \{\{\,\mid \{1\mid\,\}\}\mid\,\}\}\mid\,\}\}\\&=\{\,\mid \{\{\,\mid \{\{\,\mid 2\}\mid\,\}\}\mid\,\}\}\\&=\{\,\mid \{\{\,\mid \{0\mid\,\}\}\mid\,\}\}\\&=\{\,\mid \{\{\,\mid 1\}\mid\,\}\}\\&=\{\,\mid \{0\mid\,\}\}\\&=\{\,\mid 1\}\\&=0\end{align}$$
Claim
In general, the numeric value of $g(n)$ appears to be the highest exponent $m$ such that $2^m$ divides $n$.
Proof
Let's assume, for induction, that the claim is true of all the values of $g$ that arise during the calculation of $g(n)$. Note that $g(1)=0$. If $n$ is even, then $g(n)=\{g(n/2)\mid\,\}=g(n/2)+1$, which matches the highest power of $2$ for $n$. If $n$ is odd and greater than $1$, then $g(n)=\{\,\mid g(3n+1)\}$. Since $3n+1$ is even, $g(3n+1)\ge1$, so that $g(n)=0$, as desired.