What is the relationship between surreal star and the empty set? In Surreal Numbers and Games on page 6 it says that $0=\{\emptyset|\emptyset\}$.
Additionally, on page 10 it says that $*+*=0$ and that for any value $x$ we have $x+*=\{x|x\}$.
Given the previous, it seems logical that both $*+*=\{*|*\}=0$ and $\emptyset+*=\{\emptyset|\emptyset\}=0$ are true. 
This looks to me like $*$ and $\emptyset$ could be equivalent because given $x+*=0$, $*$ and $\emptyset$ both seem like valid solutions for $x$.
However I have not read anywhere that $*$ and $\emptyset$ are equivalent, so I am not sure if they are or if they are not.
Any clarification would be greatly appreciated.
 A: 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.
A: $0$ and $\ast$ are games.  $\emptyset$ is not a game, it's a set of games.
A “game” is a pair of sets of games.  Usually we write a pair of sets as $\langle\{\ell_1, \ell_2\}, \{r_1, r_2\}\rangle$ but in combinatorial game theory we abbreviate this to $\{\ell_1, \ell_2\mid r_1, r_2\}$.
When one or both of the sets is empty we  just leave that half of the notation empty.  So for example, $1$ is a game whose left set is $\{0\}$ and whose right set is $\emptyset$.  The usual notation would be $\langle \{0\}, \emptyset\rangle$ but in abbreviated form it is $\{ 0\mid \,\}$.
Representation of games in this way is not unique.  Two different pairs of sets might represent equivalent games.  You're right that $\ast+\ast = \{\ast\mid\ast\}$.  But the $\ast$s on the left and right turn out not to affect the value, and $\{\ast\mid\ast\} = \{\,\mid\,\} = 0$.  You can see why this should be if you consider  some game $G$ where one player has a winning strategy.  Do they also have a winning strategy in $G + \{\ast\mid\ast\}$?  Definitely.  (You work out the details.)  In the jargon, we say that the extra $\ast$ options in $\{\ast\mid\ast\}$ can be ignored because they are “reversible”.
The expression $\emptyset+\ast$ is what is called an “abuse of notation”. It is using the $+$ symbol in a way that is similar to, but not the same as, its normal meaning.  Normally, $+$ operates on two games and yields a new game, the sum.  But in expressions like $$x+y = \{X_L + y, x+Y_L\mid
X_R+y, x+Y_R\}$$ we use the $+$ in a slightly different way.  The $x+Y_L$ is a shorthand for the set consisting of everything that you can get by adding $x$ to some member of the set $Y_L$.  If $Y_L$ is $\emptyset$, then $x+Y_L$ is also empty, because there is nothing of that form.
The comma actually denotes the union of the two sets $X_L+y$ and $x+Y_L$.
We would normally write “$x+Y_L$” as “$\{x + \ell \mid \ell \in Y_L\}$”, so you can see why we need the abbreviation! 
The book does not say that “for any value $x$ we have $x+\ast = \{x\mid x\}$”.  It says that is true for any number $x$.  In general, for games that are not numbers, it is not true.  One counterexample is a game called $\ast2 $ for which $\ast2 + \ast \ne0$ but $\{\ast2 \mid \ast2\} = 0$.
