Yes, I believe there are.

I believe I came across such system in my research.

First you take the zero out of natural numbers and try to work with the remaining numbers
in a system where no number is left unrepresented. To do so you do a displacement: 1 represents 0, 2 represents 1 and so on.

You can check the introduction of the idea here (it is just the introduction):
"
https://repositorioaberto.uab.pt/bitstream/10400.2/1292/1/p_50_58.%20pdf.pdf
"

In such system, addition is different because 1 is the neutral element of addition:

2++1=2

a++1=a

And generally

a++b=a+b-1

Multiplication is therefore not commutative, because for example

2**3=3++3=5

3**2=2++2++2=(2++2)++2=3++2=4

And generally

a**b=a*b-(a-1)

But prime numbers remain the structure of this system

The new primes are simply

new prime=old prime+1

Since there is a correspondence between normal arithmetics and this system (which I have checked), the prime structure holds

There is no way to decompose a new prime other than right multiplication by two of its natural antecessor:

For example the new prime 8 can be obtained by

7**2=8 But multiplication by two on the right is just a sort of application of the neutral element of mutiplication displaced for the new system,
the equivalent of adding one unit: 7++2=8

This should not therefore be considered a "legal" multiplication for the purpose of prime identification.

Therefore primes remain primes in this non-commutative arithmetic.

I am sorry if this does not sound very rigorous. Later I can eventually provide more details and proofs. Now I am finishing a project on my daytime job.

onlymultiplication, but no addition. But then the question is as unclear as before: (1) How do you formulate induction using multiplication? The usual statement requires successor. (2) What are other axioms of the theory besides induction? The usual axioms of Robinson’s arithmetic are not applicable, as they explain multiplication in terms of addition. (3) If successor is included in the language: in standard natural numbers, $+$ is definable in terms of $\cdot$ and successor (if $c\ne0$, then $a+b=c$ if and only if $S(ac)S(bc)=S(S(ab)c^2)$). ... – Emil Jeřábek Mar 27 '13 at 13:08