To expand a bit on my comment:
In set theory one often runs into troubles when defining an equivalence relation whose universe is a class. It is quite possible to have an equivalence class (or all) that have a proper-class many members in it (e.g., all singletons).
The common trick around it is known as Scott's trick, invented by Dana Scott to allow an internal definition of cardinality in models of ZF:
Suppose that $\varphi(x,y)$ describes the equivalence relation (that is, $x$ is equivalent to $y$ if and only if $\varphi(x,y)$ is true), we then define the equivalence class of $x$ as:
$$[x]_\varphi = \{y\mid \varphi(x,y)\land\mathrm{rank}(y)\text{ is minimal}\}$$
We make a heavy use of the axiom of regularity, from which follows that every set has a von Neumann rank. We take the collection of those with minimal rank.
Note that $[x]_\varphi$ is not empty. $\varphi(x,x)$ is always true so there is some $y\in[x]_\varphi$ whose rank is at most the rank of $x$, and therefore by the replacement/subset schema $[x]_\varphi$ is indeed a set.
A minor remark about the existence of such $\varphi$ is that in ZF [proper] classes are described by a formula. If $R$ is a proper class which is also an equivalence relation then there is $\varphi$ such that $\varphi(x,y)$ is true if and only if $\langle x,y \rangle\in R$.
Notation for sets and numbers:
This is not a big problem now, but will be when you do a bit more with the surreal numbers: We must be very careful to distinguish numbers from sets of numbers.
As you said, a number is a pair of sets of numbers. That means that $(0,1)$ should definitely not be a number, since $0$ and $1$ are not sets of numbers, they're numbers. When you wrote $(0,1)$, I assume you intended $(\{0\},\{1\})$.
Similarly, When you see capital letters as in $X_L$, we are discussing sets of surreal numbers, not individual numbers. When we are clear about this, that gives us a chance to use convenient shorthands. (Here I assume you are comfortable with $\forall$ and $\in$ for "for all".) Make sure you are clear that $x\ngeq Y_R$ is a shorthand for something like $\forall y_2\in Y_R,\,x\ngeq y_2$. And $X_L\ngeq X_R$ is a shorthand for something like $\forall x_1\in X_L,\,\forall x_2\in X_R,\,x_1\ngeq x_2$.
Because of these shorthands, writing something like $X_L\ne Y_L$ would almost never be done because of the ambiguity. The reader would have to guess whether it means they are not literally the same sets (as was intended in the question post) or if it means that $\forall x_1\in X_L,\,\forall y_1\in Y_L,\,x_1\ne y_2$? etc.
Main answer:
To make this more self-contained and to clear up any confusion about the basic facts, I'll build up to $(\{0\},\{1\})$ being a number from the beginning.
Consider $(\varnothing,\varnothing)$. If we want to check if it's a number, we must check the condition $\varnothing\ngeq\varnothing$. That shorthand expands to something like $\forall x_1,x_2\in\varnothing,\ldots$, which is vacuously true. So $(\varnothing,\varnothing)$ is a number, and we call it "$0$".
Now consider $(\{0\},\varnothing)$. We must check $\{0\}\ngeq\varnothing$. But this expands to something of the form $\forall x_1\in\{0\},\forall x_2\in \varnothing,\ldots$, and is still vacuously true. So $(\{0\},\varnothing)$ is a number, and we call it "$1$".
Now consider $(\{0\},\{1\})$. Let us check whether it is a number, without knowing what background facts might help. The condition is $\{0\}\ngeq\{1\}$. That is shorthand for $\forall x_1\in\{0\},\,\forall x_2\in\{1\},\,x_1\ngeq x_2$. Since those sets only have one element each, the only statement we need to check is $0\ngeq1$. By standard convention for the slash $/$, this means we must check that $0\ge1$ is false. By standard convention for symbols with left-right orientation like this, we must check that $1\le0$ is false.
By definition, $1\le0$ (i.e. $(\{0\},\varnothing)\le(\varnothing,\varnothing)$ would mean both $1\ngeq\varnothing$ and $\{0\}\ngeq0$ hold. So we can show it is false by showing $\{0\}\ngeq0$ is false (incidentally, $1\ngeq\varnothing$ is vacuously true). And $\{0\}\ngeq0$ is shorthand for $\forall x_1\in\{0\},\,x_1\ngeq 0$. Since the set has only one element, this reduces to $0\ngeq 0 $. And to show $0\ngeq 0$ is false, we just need to show that $0\geq 0$, i.e. $0\leq 0$.
Why is $0\leq 0$ true? Well it means $0\ngeq\varnothing$ and $\varnothing\ngeq 0$. But both of those are vacuously true!
This was a lot of drilling down into definitions, so I'll reverse the order of the deductions to summarize:
- $0\ngeq\varnothing$ and $\varnothing\ngeq0$ are vacuously true.
- $0\le0$ (is true).
- $0\ngeq0$ is false.
- $\{0\}\ngeq0$ is false.
- $1\le0$ is false.
- $0\ngeq1$ is true.
- $\{0\}\ngeq\{1\}$ (is true).
- $(\{0\},\{1\})$ is a number.
On Notation:
Many sources discussing surreal numbers will use a more compact notation. $(\{a,b\},\{c,d,e\})$ would instead be written $\{a,b\mid c,d,e\}$, and something like $1=(\{0\},\varnothing)$ would be written $\{0\mid \,\}$.
Best Answer
Two different surreal numbers $x,y$ are equal if $x\leq y$ and $y\leq x$, where that order is defined in terms of their recursive parts. So you can have $$2=\{1\mid\}=\{0,1\mid\}=\{1\mid4\}=\{-17,1.5\mid\pi\}$$
It's good to think of Dedekind cuts only as far as "Ah, we can use sets of relatively simple numbers to build more complex numbers", but not a lot further than that.