The way to do this, is to calculate the coefficients of the iterates ${^2}(e^x)$, ${^3}(e^x)$, ${^4}(e^x)$, etc. first. The coefficients of the corresponding Piseaux for ${^n}z$ are going to be the same.
The problem then becomes:
Let a series $s$ with coefficients $a_n$ be given, as $s=\sum\limits_{n=0}^\infty a_n$. Determine the coefficients $b_n$ of the series expansion for the series $s_1=\exp(s\cdot x)$, where $s_1=\sum\limits_{n=0}^\infty b_n$.
There were two solutions given, one by Leroy Quet, as:
If $B(x)=\exp(A(x))$ then $B'(x)=B(x)\cdot A'(x)$, which provides a recursive relation for the solution and one by Robert Israel, as:
$$b_n=\sum\left(\prod\frac{a_{k-1}^{s_k}}{s_k!}\right)$$
where the sum is over all sequences $s=[s_1,s_2,s_3,...]$ of nonnegative integers, with:
$$n=\sum\limits_{k=1}^\infty k\cdot s_k$$
and the product is over those $k$ for which $s_k>0$ (a finite set).
(Quoting RI:) For example, the partitions of 4 correspond to $s=[4,0...]$, $[2,1,0...]$, $[0,2,0...]$, $[1,0,1,0...]$ and $[0,0,0,1,0...]$, so:
$$b_4 = \frac{a_0^4}{4!} + \frac{a_0^2 a_1}{2!1!} + \frac{a_1^2}{2!} + a_0\cdot a_2 + a_3$$
Robert then gives the following Maple code for the coefficients:
b:=(n::nonnegint)->add( mul(a[q[1]-1]^q[2]/q[2]!,
q = map(j -> [j, numboccur(j,P)], convert(P,set))),
P = combinat[partition](n));
Note OP: The method I used was Leroy Quet's, to whom I give the credits on my PhD thesis.
Note GH: If you use either method, you will end up with the coefficients of your "Pascal" Matrices.
Addendum #1:
In view of your comment, a clarification: The notation for the problem was probably a bit unfortunate. What you have, to start with, are the coefficients $a_n$ of:
$${^1}(e^x)=s=\sum\limits_{n=0}^\infty a_n\cdot x^n=\sum\limits_{n=0}^\infty \frac{x^n}{n!}$$.
The process is now inductive: You are looking for the $b_n$ such that:
$${^2}(e^x)=s=\sum\limits_{n=0}^\infty b_n\cdot x^n$$.
These $b_n$ can be gotten using either of the two solutions above. I.e. either using:
$$B(x)=\exp(A(x))\Rightarrow B'(x)=B(x)\cdot A'(x)$$
by writing down what you see and adjusting the indexes, or using Robert Israel's argument for the $b_n$.
Then you continue. You have the new $a_n$ (now as coefficients of ${^2}(e^x)$) and you are looking for new $b_n$ such that:
$${^3}(e^x)=s=\sum\limits_{n=0}^\infty b_n\cdot x^n$$.
and you continue in a similar spirit, using either method.
Addendum #2:
Here is some Maple code using R.I.'s routine, illustrating the above iteration:
restart;
N:=5;
a := Array(0 .. N, [seq(1/factorial(n), n = 0 .. N)]);# start with exp()
b := Array(0 .. N);#store coefficients of new iterate.
c := proc (n::nonnegint) options operator, arrow;
add(mul(a[q[1]-1]^q[2]/factorial(q[2]), q = map(proc (j) options operator,
arrow; [j, numboccur(j, P)] end proc, convert(P, set))),
P = combinat[partition](n)) end proc;#R.I.'s routine
Calculate coefficients of ${^2}(e^x)$:
for n from 0 to N do b[n] := c(n) end do;
Print coefficients of ${^2}(e^x)$:
for n from 0 to N do print(b[n]) end do
$$1,1,\frac{3}{2},\frac{5}{3},\frac{41}{24},\frac{49}{30}$$
Load new coefficients into array a again:
for n from 0 to N do a[n] := b[n] end do
Calculate coefficients of ${^3}(e^x)$:
for n from 0 to N do b[n] := c(n) end do
Print coefficients of ${^3}(e^x)$:
for n from 0 to N do print(b[n]) end do
$$1,1,\frac{3}{2},\frac{8}{3},\frac{101}{24},\frac{63}{10}$$
etc.
In other words:
$$\begin{align}
{^1}(e^x)&=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots\\
{^2}(e^x)&=1+x+\frac{3}{2}\cdot x^2+\frac{5}{3}\cdot x^3+\frac{41}{24}\cdot x^4+\cdots\\
{^3}(e^x)&=1+x+\frac{3}{2}\cdot x^2+\frac{8}{3}\cdot x^3+\frac{101}{24}\cdot x^4+\cdots\\
\ldots
\end{align}$$
Now, once you have the iterates of $\exp$, the corresponding Puiseux series can be gotten through the substitution $x\to \ln(z)$ by choosing the principal branch of $\ln$:
$$\begin{align}
{^1}z&=1+\ln(z)+\frac{1}{2!}\cdot \ln(z)^2+\frac{1}{3!}\cdot \ln(z)^3+\frac{1}{4!}\cdot \ln(z)^4+\cdots\\
{^2}z&=1+\ln(z)+\frac{3}{2}\cdot \ln(z)^2+\frac{5}{3}\cdot \ln(z)^3+\frac{41}{24}\cdot \ln(z)^4+\cdots\\
{^3}z&=1+\ln(z)+\frac{3}{2}\cdot \ln(z)^2+\frac{8}{3}\cdot \ln(z)^3+\frac{101}{24}\cdot \ln(z)^4+\cdots\\
\ldots
\end{align}$$
Edit#1 (22/5/2021)
By the way, I don't know why Andrew Robbins (or Wikipedia ?) has called these "Piseaux" series. These are plain (compositions of) logarithmic series, nothing less nothing more.
If you re-expand all the $\ln(z)$ terms in terms of $z$ within these, you get (examples of) Vassili Nestoridis' Universal Taylor Series. See V. Nestorides et al. on "Universal Taylor Series", etc.
