It's probably easiest to expand the exponential in the exponent first, since that will lead to a finite number of terms to be evaluated:
$$\begin{align}e^{(e^x-1)} &= \exp\left(x+\frac{x^2}{2}+\frac{x^3}{6}+O(x^4)\right)\\
&=1+\left(x+\frac{x^2}{2}+\frac{x^3}{6}+O(x^4)\right) +\frac{1}{2}\left(x+\frac{x^2}{2}+\frac{x^3}{6}+O(x^4)\right)^2
+\frac{1}{6}\left(x+\frac{x^2}{2}+\frac{x^3}{6}+O(x^4)\right)^3+O(x^4)\\
&=1+\left(x+\frac{x^2}{2}+\frac{x^3}{6}+O(x^4)\right)+\frac{1}{2}\left(x^2+2(x)\left(\frac{x^2}{2}\right)+O(x^4)\right)+\frac{1}{6}\left(x^3+O(x^4)\right)\\
&=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^2}{2}+\frac{x^3}{2}+\frac{x^3}{6}+O(x^4)\\
&=1+x+x^2+\frac{5x^3}{6}+O(x^4)
\end{align}$$
From which the coefficients can be read off straightforwardly.
$
\newcommand\c[1]{\mathcal{#1}}
$There is a good reason why we use OGF's for unlabeled structures and EGF's for labeled structures. The reason is that the multiplication for OGF's corresponds exactly the to the product for unlabeled structures, while the multiplication for EGF's corresponds to the product for labeled structures.
Unlabeled products. Let $(\mathcal A,|\cdot|_{\mathcal A})$ and $(\mathcal B,|\cdot|_{\mathcal B})$ be two unlabeled structures. The cartesian product $\c A\times \c B$ is also an unlabeled structure, with size function
$$
|(A,B)|_{\c A\times \c B}=|A|_{\c A}+|B|_{\c B}
$$
Let $a_n,b_n,c_n$ be the size sequences for $\c A,\c B,$ and $\c A\times \c B$, respectively. That is, $a_n=|\{A\in \c A\mid |A|_{\c A}=n\}|$, and analogously for $b_n$ and $c_n$. A consequence of these definitions is
$$
c_n=\sum_{k=0}^n a_k b_{n-k}
$$
Letting $A(x)=\sum_{n\ge 0}a_nx^n$ be the OGF for $\c A$, and similarly for $B(x)$ and $C(x)$, this quickly implies that $C(x)=A(x)\cdot B(x)$. Therefore, the OGF of the product is the product of the OGF's, which is a beautiful and fruitful correspondence.
Labeled products. Recall that a labeled structure is described by a function $\c A$, which takes in a finite set $X$ and returns a different finite set $\c A_X$. For example: the labeled structure of permutations is described by $X\mapsto \{\text{bijections on $X$}\}$. The only axioms required are $|X|=|Y|\implies |\c A_X|=|\c A_Y|$, and $X\neq Y\implies \c A_X \cap \c A_Y=\varnothing$.
There is also a natural notion of product $\c A\otimes \c B$ of two labeled structures. To each set $Z$, the set of $(\c A\otimes \c B)$-structures on $Z$ is defined to be
$$
(\c A\otimes \c B)_Z=\bigcup_{X\sqcup Y=Z} \c A_X \times \c B_Y
$$
The meaning of $X\sqcup Y=Z$ in the subscript is that $(X,Y)$ ranges over all pairs of sets which are disjoint and whose union is $Z$. In words, to choose the product structure, you partition $Z$ into two parts, then put an $\c A$-structure on the first, then independently put a $\c B$-structure on the second. This product really is natural; for example the labeled structure "permutations with two cycles" is the product of "cyclic permutations" with itself.
A consequence of this definition is that if $a_n,b_n,c_n$ are the size enumerators for $\c A, \c B,$ and $\c A\otimes \c B$, meaning that $a_n=|\c A_{\{1,\dots,n\}}|$ and similarly for $b_n$ and $c_n$, then
$$
c_n=\sum_{k=0}^n \binom{n}k a_k b_{n-k}
$$
This further implies that if $A(x)=\sum_{n\ge 0} a_n \frac{x^n}{n!}$ is the EGF for $\c A$, and $B(x)$ and $C(x)$ are the EGF's for $\c B$ and $\c A\otimes \c B$, then $C(x)=A(x)\cdot B(x)$. Again, the EGF of the product of labeled structures is the product of their EGF's.
Best Answer
Last fall I took a course on analytic combinatorics at my school where we covered both OGF's and EGF's extensively. The main difference between the two is a natural one. OGF's are very good at handling combinatorics of unlabelled objects. EGF's are the choice when dealing with objects in the labelled universe. This is mostly because if you have $n$ objects, you can permute their labels in $n!$ ways if there are no other restrictions.
Also, there is a practical reason that I think Qiaochu talked about. If the number of objects $a_n$ of size $n$ grows like $n!$ times other factors, using an EGF clears the $n!$. In fact, suppose that you have a sequence $\{c_n\}_{n=0}^{\infty}$ of "nice" numbers. Then, for a sequence $\{a_n\}_{n=0}^{\infty}$ you can define the generating function (maybe we can call it an AGF)
$$A(x) = \sum_{n=0}^{\infty} a_n \frac{x^n}{c_n}$$
which has the nice property that if $a_n = c_n$ for all $n$, then
$$A(x) = \frac{1}{1-x}.$$
As long as a function like $C(x) = \sum_{n=0}^{\infty} x^n/c_n$ has some nice properties, I think it would be worthwhile to use it, especially if your generating function involves objects that grow similarly to $c_n$ and you want to make a comparison. At least, this is the way I like to think about them.