How this expression leads to the given sequence
Stirling numbers of the second type satisfy
$$ \newcommand{\stirling}[2]{\genfrac\{\}{0pt}{}{#1}{#2}}\stirling{n}{0} = 0, \qquad (n \neq 0)$$
and so there is no need to define $A(0;x)$, and so the formula seems to work. You can also play with the Mathematica code which can be found in the entry corresponding to the sequence. I am puzzled, however, what is meant by "fractional iteration of $e^x-1$".

The recurrence given is for generating functions. Let's rewrite it:
$$
A_1(x) = 1, \qquad A_n(x) = \sum_{k=1}^{n-1} \stirling{n}{k} A_k(x) x.
$$
Using a table for the Stirling numbers, we get
$$
\begin{align*}
A_1(x) &amp;= 1 \\
A_2(x) &amp;= \stirling{2}{1} A_1(x) x = 1 \cdot 1 \cdot x = x \\
A_3(x) &amp;= \stirling{3}{1} A_1(x) x + \stirling{3}{2} A_2(x) x = 1 \cdot 1 \cdot x + 3 \cdot x \cdot x = x + 3x^2 \\
A_4(x) &amp;= \stirling{4}{1} A_1(x) x + \stirling{4}{2} A_2(x) x + \stirling{4}{3} A_3(x) x \\ &amp;= 1 \cdot 1 \cdot x + 7 \cdot x \cdot x + 6 (x + 3x^2) x  = x + 13x^2 + 18x^3
\end{align*}
$$
and so on. The numbers in the table are $a(n,k) = A_n(x)|_{x^k}$ :
$$
\begin{align*}
a(2,1) &amp;= 1, \\
a(3,1) &amp;= 1, \\
a(3,2) &amp;= 3, \\
a(4,1) &amp;= 1, \\
a(4,2) &amp;= 13, \\
a(4,3) &amp;= 18,
\end{align*}
$$

If you prefer an actual recurrence on the coefficients, here is what you get:
$$
\begin{align*}
a(n,l) = \begin{cases} 1 &amp; l = 1, \\ \sum_{k=l}^{n-1} \stirling{n}{k} a(k,l-1) &amp; l &gt; 1. \end{cases}
\end{align*}
$$