Does there exist a set $A\subset\mathbb{N}$ such that, for each $d\in\mathbb{N}$ there is a unique $n\in\mathbb{N}$ such that $n\in A$ and $n+d\in A?$ 
Does there exist a set $A\subset \mathbb{N}$ such that, for each
$d\in\mathbb{N},\ $ there is a unique $n\in\mathbb{N}$ such that $n\in
A$ and $n+d \in A\ ?$

My first thought was that the set of triangular numbers, $T,$ settles this. However, upon inspection, we see that $T$ fails for $d=5$ due to each of $\ 1,6,10\ $ and $15\ $ being members of $\ T.$
I tried to find a set with the greedy algorithm, that is, starting with $A=\{\},\ $ append to $A$ the smallest positive integer that doesn't itself violate the condition in the question, and repeat this process. Or to be more precise about this process, append to the current iteration of $A$ the smallest positive integer $x$ such that $\{\vert x - y \vert: y \in A \}\ \cap \{\vert a - b \vert: a,b \in A \}\ =\emptyset.$
Doing this gives us the set:
$$A = \{ 1, 2, 4, 8, 13, 21, 31, 45, 60, 76, 97, 119, 144, 170, 198, 231, 265, 300, 336, 374, 414, 456, 502, 550, 599, 649, 702, 759, 819, 881, 945, 1010, 1080, 1157, 1237, 1318, 1401, 1486, 1572, 1662, 1753, 1845, 1945, 2049, 2156, 2264,\ldots \}.$$
which is https://oeis.org/A004978.
However, it turns out that this set fails to meet the condition in the question for $d=110$ due to each of $60,170,$ and $649, 759$ being members.
So the question stands...

We can do this greedily, but instead of adding in one number at a time, we'll add in two numbers at a time.
Suppose that we've built the sequence $a_0, a_1, a_2, a_3, \dots, a_{2k}, a_{2k+1}$. Let $$D_k = \{a_j - a_i : 0 \le i < j \le 2k+1\}$$ be the set of all differences that exist between the terms of this sequence.
First, let $d$ be the smallest natural number not in $D_k$. Then, let $a$ be the first natural number after $a_{2k+1}$ such that the difference sets
$$\{a - a_i : 0 \le i \le 2k+1\} \text{ and } \{(a+d) - a_i : 0 \le i \le 2k+1\}$$ have no overlap with $D_k \cup \{d\}$. These difference sets also can't overlap with each other: if $a - a_i = (a+d)-a_j$, then we get $d = a_j - a_i$, contradicting the assumption that $d \notin D_k$.
By construction, if we set $a_{2k+2}$ and $a_{2k+3}$ to $a$ and $a+d$, no repeat differences are created. On the other hand, every difference is eventually created once as $(a+d)-a$ if it is not created earlier.
If my code is correct, the greedy sequence we get begins
$$1, 2, 5, 7, 15, 22, 38, 47, 65, 76, 120, 132, 154, 173, 241, 265, 
327, 353, 482, 510 \dots$$
Another nearly-greedy solution with this property is OEIS A111328, which takes nearly the same strategy, but with the specific choice $a_{2k+2} = 2a_{2k+1}+1$; this is guaranteed to be far enough to avoid repeated differences, though it is not always the smallest choice.
