Yes. Denote $X_k=(k, f(k))\in \newcommand{\Nplus} {\mathbb{N}^+}\Nplus\times \Nplus$. We want to construct the point set $A:=\{X_1,X_2,\ldots\}\subset \Nplus\times \Nplus $ so that:

1. $A$ contains exactly one point on each vertical line $\{i\}\times \Nplus$;

2. $A$ is a Sidon set: all difference between elements of $A$ are distinct;

3. every vector $(a, b) \in \Nplus\times \Nplus$ is realized as a difference of two elements of $A$.

This may be done by a standard procedure with adding points to $A$. On $(2k-1)$-th step you add a point on the $k$-th vertical line if it still does not exist; on $2k$-th step you add two points whose difference is the $k$-th vector from the set $\Nplus\times \Nplus$ (enumerated in advance) if there are still no two such points; and always care on property 2.

Yes. Denote $X_k=(k, f(k))\in \newcommand{\Nplus} {\mathbb{N}^+}\Nplus\times \Nplus$. We want to construct the point set $A:=\{X_1,X_2,\ldots\}\subset \Nplus\times \Nplus $ so that:

1. $A$ contains exactly one point on each vertical line $\{i\}\times \Nplus$;

2. $A$ is a Sidon set: all differences between elements of $A$ are distinct;

3. every vector $(a, b) \in \Nplus\times \Nplus$ is realized as a difference of two elements of $A$.

This may be done by a standard procedure with adding points to $A$. On $(2k-1)$-th step you add a point on the $k$-th vertical line if it still does not exist; on $2k$-th step you add two points whose difference is the $k$-th vector from the set $\Nplus\times \Nplus$ (enumerated in advance) if there are still no two such points; and always care on property 2.