Does anything precede incrementation in the operator "hierarchy?"

We can notice that

$a^{S(n)}=mul_a(a^n)=a\times (a^n)$

$a \times S(n)=add_a(a\times n)=a+(a\times n)$

$a+S(n)=S(a+n)$

here the seqence breaks, since successor is not a binary operation. but we can continue finding a function $f$ that is the "$(-1)$ step of the sequence (if the Successor is the step $0$).

$S(S(n))=f(S(n))$ that becomes

$n+2=f(n+1)$ and for $n=m-1$ we get

$(m-1)+2=f(m)=m+1$

so $S(n)=f(n)$: "successor precedes" the successsor in the sequence

To expand on MphLee's answer:

Successor can also be viewed as a binary function $H_0(a,b) = S(b)$. Then we have that $H_0(a, Sb) = H_{-1}(a,H_0(a,b)) = H_{-1}(a,Sb)$ so $H_{-1}$ agrees with $H_0$ when the second argument is greater than $0$. However, if we stop there, $H_{-1}(a,0)$ can be specified freely.

If we continue further, we can show that $H_{-k}(a, b) = S(b)$ for $b \ge k$:

If $b \ge k$, then $SSb = H_{-k}(a, Sb) = H_{-k-1}(a, H_{-k}(a,b)) = H_{-k-1}(a, Sb)$.

However, then $2 = H_{-1}(a, 1) = H_{-2}(a, H_{-1}(a,0))$, so if $f(a) := H_{-1}(a,0) \ge 2$, we get that $2 = f(a) + 1$, so $f(a) = 1$, a contradiction. Therefore, $f(a) \le 1$.

If we continue further it gets more complicated. I think if you extend it infinitely, they all have to be successor, even though they don't in a finite extension.