This is not a complete answer
This relation can be formalized as definition by recursion:
if we can create an infinite sequence of binary operation on $\Bbb N$ in this way:
First we define $+_1:=+$
we can define $+_{k+1}$ from $+_{k}$ using the recursion over naturals with three steps.
$i)\,\,\,\, m +_{k+1} 0= b_{k+1}$
$ii)\,\,\,\, m +_{k+1}(n+1)=m+_k (m +_{k+1}n)$
$iii)\,\,\,\, b_ {n+1}:= \begin{cases} b, & \text{if $n=0$} \\ 0, &
> \text{if $n=1$ } \\ 1, & \text{if $n\gt 1$ } \\ \end{cases}$
In this way we can go from $+_{k}$ to $+_{k+1}$.
This is the same way we build the Hyperoperations (see here)
$i)\,\,\,\, H_0(a,b):=S(b)=b+1$
$ii)\,\,\,\,H_{n+1}(a,0):=a_{n+1}$
$iii)\,\,\,\,H_{n+1}(a,b+1):=H_n(a,H_{n+1}(a,b))$
$iv)\,\,\,\, a_ {n+1}:= \begin{cases} a, & \text{if $n=0$} \\ 0, &
> \text{if $n=1$ } \\ 1, & \text{if $n\gt 1$ } \\ \end{cases}$
As we can see your $\epsilon$ has this property
$$\epsilon (+_{k})=+_{k+1}$$
About such $\epsilon$ that in our case it takes a binary operation as a imput and give a binary operation as output but I never saw it. Instead I saw a similar concept for unary fucntions called Superfunction.
We call $F$ superfunction (related with Abel Functions) the solution of the functional equation
$$F(0)=b$$
$$F(n+1)=h(F(n))$$
Q I don't know if there is a notation for this solution but I am interested to know if there is a standard one.
I usually write $F$, the solution of this equation, as $\mathcal S [h]$:
$$F=\mathcal S [h]$$
$$ \mathcal S [h](n+1)=h(\mathcal S [h](n)) $$
because I saw it somewhere but I don't remember where.
So if we use this notation for our operations turning them in to unary functions for a fixed $b$
$f_0(n):=H_0(b,n)=n+1$
$f_k(n):= H_k(b,n)=b +_{k} n$
we can find that $f_{k+1}(n):= H_{k+1}(b,n)=b +_{k+1} n$ is the Superfunction of $f_{k}(n)$
so where $f_1(n)=b +_{1} n=b+ n $ we have
$$\mathcal S [f_1](n)=f_2(n)=b +_{2} n=b\times n $$
and in general
$$\mathcal S [f_{k}]=f_{k+1} $$