*[update]* Here -for possibly a nicer reference- I give a screenshot, where the identity by the OEIS-formulae (and kindly expanded by @Carlo Beenakker) is interpreted as matrix-multiplication. It indicates also, that the given formula of relations between the Stirling and A-numbers is near to an eigenvector/eigenmatrix-relation (in fact is a Jordan-decomposition) which I described in my first version of this post (which I kept below).

Here is the matrix-multiplication scheme:

*[the original answer: Jordan-decomposition]*

Another approach gave the numbers in a completely surprising context; I've no explanation so far. *([update]:It is clear now. Carlo's expansion gave the key hint!)*.

For the function $\exp(x)-1$ the matrix $S_2$ of Stirling numbers 2'nd kind is involved. I've just recently begun to exercise with the Jordan-decomposition, and for some exercise I took this matrix

$$ \large S_2 \qquad = \qquad \small{ \begin{bmatrix}
1 & . & . & . & . \\
1 & 1 & . & . & . \\
1 & 3 & 1 & . & . \\
1 & 7 & 6 & 1 & . \\
1 & 15 & 25 & 10 & 1
\end{bmatrix}} $$
and let WolframAlpha Jordan-decompose it such that $$S_2 = A \cdot J \cdot A^{-1}$$

This gave the three matrices:

$$ \begin{array} {} A &=&\small { \begin{bmatrix}
1 & . & . & . & . \\
0 & 1 & . & . & . \\
0 & 1 & 3 & . & . \\
0 & 1 & 13 & 18 & . \\
0 & 1 & 50 & 205 & 180
\end{bmatrix} } \\
J&=& \small {\begin{bmatrix}
1 & . & . & . & . \\
1 & 1 & . & . & . \\
0 & 1 & 1 & . & . \\
0 & 0 & 1 & 1 & . \\
0 & 0 & 0 & 1 & 1
\end{bmatrix} } \\ A^{-1}&=&\small {\begin{bmatrix}
1 & . & . & . & . \\
0 & 1 & . & . & . \\
0 & -1/3 & 1/3 & . & . \\
0 & 5/27 & -13/54 & 1/18 & . \\
0 & -301/2430 & 353/1944 & -41/648 & 1/180
\end{bmatrix}}\end{array}$$
(Note: I've transposed the input to Woframalpha and then also the output to keep in line with my usual conventions with that type of matrix-discussion)

The surprise is: *that we find the coefficients in the matrix $A$*

The relation can also be written as
$$ S_2 \cdot A = A \cdot J $$
and if $J$ is decomposed in the sum of two (diagonal and subdiagonal) components $J=J_0+J_1$ where $J_0=I$ and $J_1$ the transpose of my $J$ in the update of this post above, then we can reformulate
$$ S_2 \cdot A = A \cdot (I+J_1) \\
(S_2 - I) \cdot A = A \cdot J_1 \\
S_2^* \cdot A = A \cdot J_1 \\
$$
from which -in my opinion- we could formulate a slightly more convenient relation than that given in the OEIS.

Additional remark because there is a relation to the Schröder-function for fractional iteration: for the use for the function $f(x) = \exp(x)-1$ the matrix $S_2$ is factorially similarity-scaled and -when let unscaled- becomes the Bell/(transposed) Carleman-matrix for that function. In that view the scaling $F^{-1} \cdot S_2 \cdot F$ is a matrixoperator; and the matrices $A$ and $A^{-1}$ are in a very similar role as the operators for the Schröder-function of $f(x)$ which is indeed at the heart of the Schröder/Abel-type of fractional iteration...

This is the input for W/A's input field:

JordanForm({{1,1,1,1,1},{0,1,3,7,15},{0,0,1,6,25},{0,0,0,1,10},{0,0,0,0,1}})

Mathematicafunction at OEIS works as desired. $\endgroup$ – Fred Kline Jun 13 '13 at 8:36