The (un)decidability of Robinson-Arithmetic-without-Multiplication?

It is undecidable.

Let $\mathcal L$ be the first-order language with equality and one binary predicate $p({-},{-})$ and no constants or function symbols. Call an $\mathcal L$-sentence of the form $\varphi\land \exists x\exists y(x\ne y)$ "fancy" -- it is then well known that satisfiability of fancy sentences is undecidable.

Consider the following translation from $\mathcal L$ into the your additive Robinson arithmetic:

$$\overline{\varphi\land\psi} \equiv \overline\varphi\land\overline \psi \qquad \overline{\neg\varphi} \equiv \neg\overline\varphi$$ $$\overline{\exists x.\varphi} \equiv \exists x(x=\mathsf S x \land \overline\varphi) \qquad \overline{x=y} \equiv x=y$$ $$\overline{p(x,y)} \equiv x+y=x $$

Now clearly if $\varphi$ is a fancy sentence and $\overline{\varphi}$ is consistent with additive Robinson arithmetic, then $\varphi$ is satisfiable -- it is trivial to derive a model of the latter from a model of the former; the domain will be the elements that are their own successors.

On the other hand if $\varphi$ is a satisfiable fancy sentence, then there's a model of additive Robinson arithmetic that satisfies $\overline\varphi$, as follows:

Let $(A,p)$ be a model of $\varphi$; without loss of generality take $A$ to be disjoint from $\mathbb N$. Because $\varphi$ is fancy $A$ has at least two elements, so let $f:A\to A$ be such that $f(a)\ne a$ for all $a\in A$. Now construct an model of additive Robinson arithmetic with domain $\mathbb N\cup A$, as follows: $$ 0 = 0 $$ $$ \mathsf S n = (n+1) \qquad \mathsf Sa = a$$ $$ n\oplus n' = (n+n') \qquad a\oplus n= n\oplus a = a \qquad a\oplus a' = \begin{cases} a & \text{when }p(a,a') \\ f(a) & \text{otherwise} \end{cases}$$

Consider the model $\mathfrak{M}$ with domain the natural numbers $\mathbb{N}$ plus one more object, say $a$. Define $S$ on $\mathbb{N}$ normally and $Sa=a$. Define $+$ on $\mathbb{N}$ normally and $x+a=a+x=a$ for all $x \in \mathfrak{M}$.

Then it is quite easy to prove that $\mathfrak{M}$ is a model of the $5$ axioms in which the statement $\varphi$, "there exists an $x$ such that $Sx=x$" is true.

So $\varphi$ cannot be decided by the axioms since in the standard model it is false and true here, so the theory is undecidable.