For what subset of the reals is the difference of any two elements always unique such that every real can be so represented? I am looking for a subset $A$ of the real numbers such that given a real number $z$ not equal to $0$ there exists a unique $x,y \in A$ such that $x-y=z$. Or for any real number $z$ not equal to $0$, $x-y=z$ for $x,y \in A$ has one and only one solution.


Some examples of what can't happen:
If $A=\{0,1,2\}$ then if $z=1$ we have two solutions, $2-1=1$ and $1-0=1$.
If $A=\mathbb{Z} $ then $z=.5$ has no solutions.
Another way I have been looking at this problem is as points on a number line where there is always exactly one set of points any given distance apart.
EDIT: dropped largest/maximal from the question given Patrick's lemma.
 A: (Work in progress.)
Say a subset $A$ of $\mathbb{R}$ is difference-unique if for all $z \not = 0, z \in \mathbb{R}$ there is at most one pair $a_z, b_z \in A$ with $a_z - b_z \in A$. Say $A$ is nice if additionally for all $z \not = 0, z \in \mathbb{R}$ there is exactly one $a_z, b_z \in A$ with $a_z - b_z = z$. (That is, your property.) It is immediately obvious that nice sets must be uncountable, because they must have uncountably many differences-of-two-elements, and they must be unbounded, because arbitrarily large differences must be formed.
WLOG also that if $A$ is difference-unique (or nice), then $0 \in A$. Indeed, if not, we may select an element $a \in A$ and consider $A - a$, which has exactly the same difference-uniqueness/niceness status as $A$.
Lemma: Let $A$ be nice. Then $A$ is not a strict subset of another nice set.
Indeed, if $B$ were any set with $A \subset B$, then say $b \in B \setminus A$, and pick some other $c \in B$. Then there is unique $x, y \in A$ with $x-y = c-b$; but $x, y$ are in $B$ also, so $B$ is not nice.

A Zorn's lemma proof will hopefully be possible.
Lemma: the union of a nested collection of difference-unique sets is difference-unique. (That is, chains in the poset of difference-unique sets, ordered by inclusion, have upper bounds.)
Proof: suppose $\cup_I A_i$, a union of difference-unique sets, were not difference-unique. Then there would be $a, b, c, d$ in the union which had $a-b = c-d$, with $a \not = c, b \not = d$, $a \not = b$, $c \not = d$. Then all $a,b,c,d$ appear in some $A_i$, contradicting the difference-uniqueness of each $A_i$.
Therefore, by Zorn's lemma, there is a maximal difference-unique set. We just need one of these to be nice.

The obvious greedy algorithm ("find the first integer which is not a difference, add it to the last integer in our sequence, and append that sum to the sequence") fails to build an "integer-nice" $A$: a set of integers such that every integer appears exactly once as a difference. It builds $\{0, 1, 3, 7, 12, 20, 30, 44\}$ and then attempts to add $59$ to the list (while attempting to get $15$ as a difference), which has $29 = 59 - 20 = 30 - 1$.
