Does the given proof show that $0$ is the unique additive identity? 
The number $0$ has one property so important that we list it next:
$(P2)$ If a is any number, then
$ a + 0 = 0 + a = a$.
An important role is also played by $0$ in the third property of our list:
$(P3)$ For every number $a$, there is a number $-a$ such that
$a+(-a) = (-a) + a = 0$.
Property $P2$ ought to represent a distinguishing characteristic of the number $0$, and it is comforting to note that we are already in  a position to prove this. Indeed, if a number $x$ satisfies
$a+x=a$
for any one number $a$, then $x=0$ (and consequently this equation also holds for all numbers $a$). the proof of this assertion involves nothing more than subtracting $a$ from both sides of the equation, on other words,adding $-a$ to both sides; as the following detailed proof shows, all three properties $P1-P3$ must be used to justify this operation.
$(P1)$ If $a,b$, and $c$ are any numbers, then
$a + (b+c) = (a+b)+c$
proof. If $a+ x = a$,
then $(-a) + (a+x) = (-a)+a = 0$;
hence $((-a) + a) + x = 0$;
hence $0+x=0$;
hence $x=0$

From $(P2)$, we are shown that there exists an additive identity. When we prove that there exists a number $x=0$ that satisfies $a + x = a$ for any number $a$, is this the same as showing that $0$ is the unique additive identity?
 A: 
When we prove that there exists a number $x=0$ that satisfies $a + x = a$ for any number $a$, is this the same as showing that $0$ is the unique additive identity?

No. In general: an existence proof is not the same as a uniqueness proof.
(also: please note that we don't give any existence proof at all.  That is, we don't prove that there exists a number $x=0$ that satisfies $a + x = a$, but stipulate it through (P2))
However, contrary to the claim that:

all three properties $P1-P3$ must be used to justify this operation. 

you can actually prove that $0$ is the only additive identity from (P2) alone:
Suppose that $x$ is some additive identity. Then we have that $a+x=a$ for any $a$.  But since this holds for any $a$, we can use $a=0$, and thus we have that $0+x=0$. Now, by (P2) we have that $0+a=a$ for any $a$, and hence using $a=x$, we have that $0+x=x$. Combined with the earlier $0+x=0$ we thus have $x=0$
Still, please note that this proof that shows that for any $x$: if for all $a$: $a+x=a$, then $x = 0$ relies on the fact that $0+a=a$ for any $a$. So, when you suggest that the existence of the number $x=0$ that satisfies $a + x = a$ (i.e. that $a+0=a$ for all $a$) implies that $x=0$ would be the only $x$ for which  it holds that for all $a$: $a+x=a$, that is still not quite right, because what you need is that $0+a=a$ for any $a$, rather than that $a+0=a$ for all $a$. Both are part of (P2), but you pointed to the wrong part.
Then again, it may be better to define the uniqueness theorem as:
For any $x$: if for any $a$: $a+x=a$ and $x+a=a$, then $x=0$
And, defined as such, we can prove the uniqueness theorem from the mere fact that $a+0=a$ for all $a$ (You try it!)
