In the Peano arithmetic wikipedia article, a figure is shown on why the axiom of induction is necessary, without it, the set of whole dominos would be a valid representation of N
In Robinson arithmetic, we have the axiom y=0 ∨ ∃x (Sx = y) instead.
I don't see how this axiom prevent that situation, the whole set of dominos still looks valid as satisfying the axioms of Robinson arithmetic. Am I wrong ?
This image is really only about how the successor function behaves, so is more suitable to a discussion of the Peano axioms than to a discussion of Peano Arithmetic. But more to the point: in order to say that this can satisfy the axioms of Robinson arithmetic$^*$, you need to specify how addition and multiplication are defined.
That said, we can certainly get a model that has this shape. Let's say there is only one element in the circular part... call it $a.$ Of course, we have $Sa = a.$ Then for addition and multiplication we can take $a+x=x+a=a,$ and $x\cdot a = a\cdot x = a,$ for all instances involving $a$, except of course we need to keep $a\cdot 0 = 0.$ All the elements of the model that are not $a$ just behave like the natural numbers, and all arithmetic without $a$ in it remains the same. It's not difficult to see this satisfies all the axioms of Robinson arithmetic.
I don't know how to systematically extend this to longer cycles, but it's not too hard to work out addition and multiplication functions that make a model of Robinson arithmetic with extra elements $a$ and $b$ with $Sa= b$ and $Sb = a.$
Finally, would be remiss to not point out that while first-order Peano Arithmetic rules out the picture in the image you posted, it doesn't rule out nonstandard "disconnected" successor functions in general, although there will always be infinitely many components and there will never be cycles. (Also, more generally, they will never be of the sort of the simple models of Robinson arithmetic above where you can just "work out" some addition and multiplication function: all nonstandard models of PA are non-computable.) On the other hand, the second-order Peano axioms do guarantee that all models are the same as the natural numbers, up to isomorphism.
$^*$ It's fairly standard, but since there are inevitably variations (and it's a very weak theory, so can be very sensitive to seemingly minor variations), we're taking Robinson arithmetic to be the following axioms in the language $(0, S,+,\cdot):$
As an example of what slight strengthening could do, imagine we extended to include a symbol $<$ for the ordering. Say then we we included axioms saying $<$ is a total ordering, and also that $x < S x.$ Seems innocuous enough, but observe it completely changes the answer to your question since with these axioms added, there can't be any models where the successor relation goes around in a circle. But on the other hand. it doesn't rule out extra components that look like $\mathbb Z$ rather than a circle.