Proof that sets with 1 element exist from these axioms I have:


Axiom of existence: there exists a set with no elements
Axiom of extensionality (equality basically?): if two sets have the same elements they are identical
Lemma showing the empty set is unique
Axiom of schema of comprehension: $\{x\in A|P(x)\}$ exists where P is a property.
Lemma showing the set given in the comprehension axiom is unique (justifying my notation)
Axiom of a pair: for any A, B there is a set C such that $x\in C\iff[x\in A\text{ or }x\in B]$


The book then goes on to say "We introduce the notation {A,B} for the unordered pair of A and B, in particular if A=B we write {A} instead of {A,A}"

Is this suggesting that 1 element sets exist? Can I justify that with comprehension? 

Or is it simply just notation "We're lazy and don't want to write {A,A}" (I'm hoping for this)

I ask because I'm trying to be rigorous, so far I am happy that: sets containing nothing exist, I can have pairs (but I must not think sets containing two things, because {A} is just one thing) right?

I do not like that pair means "set containing two things"
This is how I like to think about it. The first remark uses the axiom of extensionality.



[Sorry for the picture; I prefer to typeset in my LateX editor with \newcommands + autocompletion]
Since $A$ in the axiom of  schema of comprehension, and $A$ and $B$ in the axiom of pair, are sets, nothing in the 5 axioms and lemmas you provided says that there exists a set other than the empty set (that is, pair axiom as you have stated it does not give you a set containing two elements).

So it is no surprise that you can't see it logically implying existence of a set with one element.