What is the series of identities for these hyper-operations? Suppose we define the hyper-operations iteratively off the previous hyper-operation. We begin with the zeroth hyperoperation being addition,
$$a\circ_0b:=a+b.$$
We then assert that
$$a\circ_{n+1} b:=\exp(\ln(a)\circ_n\ln(b)).$$
Although not immediately obvious, $a\circ_1 b\equiv a\times b$. So $\circ_1$ is equivalent to multiplication. But $a\circ_2 b\neq a^b$, but instead reduces to $a^{\ln(b)}$. Thus $\circ_2$ is not the traditional exponentiation (instead, the so-called "powerlog" in this video). Doing so allows each hyper-operation to be associative and communitive over it's previous operation, which are nice properties that regular exponentiation does not have.

My question is: What is the series of identities, $e_n$, such that $a\circ_n e_n \equiv a$, and what is the limit of $\lim_{n\to\infty}e_n$?

We know that $e_0=0$, as $a\circ_00\equiv a$, since $a\circ_0b:=a+b$.
We also know that $e_1=1$, as $a\circ_11\equiv a$, since $a\circ_1b:=a\times b$
I can also deduce that $e_2=e$, as $a\circ_2 e\equiv a$, since $a\circ_2b:=a^{\ln b}$.
But I would like to know if there is a general formula for $e_n$, and what the limit is (if any). Kind regards to any insight.
 A: By your definition:-
$a\circ_{n+1}b=\exp(\ln(a)\circ_n\ln(b))$
It is clear that $a\circ_nb=a^{\ln(\cdots(\ln(b))}$
There are $n-1$ '$\ln$'s
But still, I think I should still give some clarification-
$a\circ_0b=a+b$
$\begin{align}a\circ_1b & =\exp(\ln(a)+\ln(b))\\
&=a×b\end{align}$
$\begin{align}a\circ_2b& =\exp(\ln(a)\ln(b))\\
&=a^{\ln(b)}\end{align}$
$\begin{align}a\circ_3b &=\exp(\ln(a)^{\ln(\ln(b))})\\
&=a^{\ln(\ln(b))}\end{align}$
$$\vdots$$
$\begin{align}a\circ_nb &=\exp(\ln(a)^{\ln(\cdots(\ln(b)))})\\
&=a^{\ln(\cdots(\ln(b)))}\end{align}$
$\boxed{a\circ_nb=a^{\ln(\cdots(\ln(b)))}}$

To find the identity $e_n$
$a\circ_ne_n=a$
$a^{\ln(\cdots(\ln(e_n)))}=a$
$\ln(\cdots(\ln(e_n)))=1$
$$e_n=e^{e^{\unicode{x22F0}^{e}}}$$
It is a power tower of $(n-1)$ $e$s
$\therefore e_n=^{(n-1)}e$ for $n\geq 1$, which doesn't include the $e_0=0$ case, so putting that in, we get-
$e_n=\begin{cases}0 & \text{if}, n=0\\^{(n-1)}e & \text{if}, n\geq 1\end{cases}$
And by that definition it is clear that $\textstyle\displaystyle{\lim_{n\rightarrow\infty}e_n}$ diverges
I think that answers your question.
