Puzzle: Can arithmetic be axiomatized with a single two-term relation?

I think your result is actually a special case of ($\star$) given any (finite) number of arbitrary relations $R_1,R_2, \ldots R_r$ on $\mathbb Z$ (with arbitrary arities), there is a single two-term relation $A(x,y)$ on $\mathbb Z$ such that all the $R_i$ are definable in terms of $A$.

To prove ($\star$), it will suffice to show the weaker form ($\star\star$ ) : given any two two-term relations $R_1(x,y)$ and $R_2(x,y)$ on $\mathbb Z$, there is a two-term relation $A(x,y)$ on $\mathbb Z$ such that $R_1$ and $R_2$ are both definable in terms of $A$ (indeed, it follows from ($\star\star$) by induction that any finite number of two-terms relations can be generated by a single two-term relation, and if $\beta=(\beta_1,\beta_2)$ is any bijection ${\mathbb Z} \to {\mathbb Z}^2$, we may add the two-term relations $U_1(x,y):y=\beta_1(x)$ and $U_2(x,y):y=\beta_2(x)$. Then $\beta(z)=(x,y)$ can be defined as $U_1(z,x) \wedge U_2(z,y)$, so that $\beta$ is definable. This takes care of relations with larger arities).

So let us show ($\star\star$) : take two two-term relations $R_1(x,y)$ and $R_2(x,y)$ on $\mathbb Z$. Let $X=\lbrace 501,601,701,801, \ldots ,1401\rbrace$ and $Y=\lbrace 10;11;12;13;14 \rbrace$. Since $|X|=10$ and $|Y|=5$, we have $|X|<2^{|Y|}$ so there is an injection $\gamma : X \to {\cal P}(Y)$.

Define a two-term relation $A : {\mathbb Z}^2 \to \lbrace True,False \rbrace$ by the following eleven (disjoint and hence consistent) rules :

(1) For $x\in{\mathbb Z}$, set $A(x,x)=(x=5)$.

(2) For $x\in [|5(...)14|], y\in{\mathbb Z} \setminus \lbrace 5 \rbrace$ with $y \neq x$, set $A(x,y)=(y=x+1)$.

(3) For $x\in{\mathbb Z} \setminus [|5(...)15|], y \in [|5(...)8|]$, set $A(x,y)=( x \equiv y-4 \ ({\rm mod} \ 100))$.

(4) For $x\in{\mathbb Z} \setminus [|5(...)15|]$, set $A(x,9)=(x\in X)$.

(5) For $x\in X$ and $y\in Y$, set $A(x,y)=(y\in \gamma(x))$.

(6) For $x,y \in \mathbb Z$, set $A(100x+1,100y+2)=(x=y)$.

(7) For $x,y \in \mathbb Z$, set $A(100x+2,100y+1)=(R_1(x,y))$.

(8) For $x,y \in \mathbb Z$, set $A(100x+1,100y+3)=(x=y)$.

(9) For $x,y \in \mathbb Z$, set $A(100x+3,100y+1)=(R_2(x,y))$.

(10) For $x\not\in [|5(...)14|],y \in \mathbb Z$ with $x\neq 100y+4$, set $A(x,100y+4)=(y=x)$.

(11) For $x\in{\mathbb Z}, y\not\in [|x(...)14|], y\not\equiv 4 \ ({\rm mod} \ 100)$, set $A(100x+4,y)=(y=100x+1)$.

Then, using only $A$, we may successively define the one-term relations ‟$x=5$”, ‟$x=k$” for $2 \leq k \leq 14$, ‟x equals $k$ modulo 100” for $1\leq k \leq 4$,‟$x$ is in $X$”, and finally ‟$x=k$” for each individual $k\in X$ (using rule (5)).

Then the two-term relation $D(x,y):y=100x+1$ can be defined : this is $D_1(x,y) \vee D_2(x,y)$, where $D_1(x,y)$ is $$ (x=5 \wedge y=501)\vee(x=6 \wedge y=601) \ldots \vee (x=14 \wedge y=1401) $$ and $D_2(x,y)$ is $$ (y\not\in [| 5(...)15 |]) \wedge (y\not\equiv 4 \ ({\rm mod} \ 100)) \wedge (\exists z \ (z \equiv 4 \ ({\rm mod} \ 100)) \wedge A(x,z) \wedge A(z,y)). $$

Now we can define $R_1(x,y)$ as $$ \exists x' \ \exists y' \ \exists z \ (z \equiv 2 \ ({\rm mod} \ 100)) \wedge D(x,x') \wedge D(y,y') \wedge A(x',z) \wedge A(z,y') $$ and $R_2(x,y)$ as $$ \exists x' \ \exists y' \ \exists z \ (z \equiv 3 \ ({\rm mod} \ 100)) \wedge D(x,x') \wedge D(y,y') \wedge A(x',z) \wedge A(z,y'), $$ which finishes the proof.