Inclusion-exclusion can be thought of, in some sense, as a special case of the use of generating functions, but both should be understood in a fairly broad sense for this to be true. A broad context for inclusion-exclusion is Möbius inversion on a poset, which reduces to ordinary inclusion-exclusion in many cases such as when the poset is the poset of subsets of a set. Möbius inversion naturally takes place in the incidence algebra of a poset, which can be thought of as a place where generating functions associated to the poset naturally live. In particular, three of the most important types of generating functions (ordinary, exponential, Dirichlet) can be thought of as associated to posets in roughly this manner, although the details are a bit subtle.
For details, see Doubilet, Rota, and Stanley's On the foundations of combinatorial theory (VI): the idea of generating function.
Just to get the question clear: The descent set of a permutation $\pi$ is defined as the points where a descent occurs, namely $\operatorname{Des}(\pi) = \{i: \pi_i > \pi_{i+1}\}$.
The first descent in a permutation of size $n$ is defined as $F(\pi) = \min (\operatorname{Des}(\pi) \cup \{n\})$. (The union with $\{n\}$ is just to cover the case of the permutation with no descents, namely $(1, 2, \dots, n)$, for which define the first descent to be at the last position $n$.)
A "desarrangement" is defined as a permutation $\pi$ for which $F(\pi)$ is even.
We want to find the exponential generating function for $E_n$, the number of desarrangements of length $n$.
Method 1.
Prove that the number of desarrangements is the same as the number of derangements, via a bijection. As this is the approach Brian M. Scott alludes to in his answer, I will not elaborate on this further. (Besides, it requires you to know / derive the EGF of the number of derangements.)
Method 2.
We'll count the number of permutations $\pi$ of size $n$ that have $F(\pi) = 2k$, for each $k$.
To have $F(\pi) = 2k$, if $2k < n$, it means that the relative order of the first $2k+1$ elements is such that they are all ascents, except for a final descent at position $2k$. The probability (over uniform permutations of size $n$) that this happens is $\dfrac{2k}{(2k+1)!}$, as of the $(2k+1)!$ possible orderings of the first $(2k+1)$ elements, only $2k$ have this property. (Essentially, arrange all $(2k+1)$ of them in increasing order, and then move any of the first $2k$ non-maximum elements to the end.)
The other way in which we can have $F(\pi) = 2k$, per our definition is if $n = 2k$ and the permutation is $(1, 2, \dots, n)$.
So the number of permutations of size $n$ with $F(\pi)$ even is
$$E_n = \sum_{0 \le 2k < n} n!\frac{2k}{(2k+1)!} + [n\text{ is even}]$$
The EGF is
$$
\begin{align}
E(z) = \sum_{n\ge 0} E_n \frac{z^n}{n!}
&= \sum_{n\ge0}\sum_{0\le 2k < n} n!\frac{2k}{(2k+1)!} \frac{z^n}{n!} + \sum_{n \ge 0} [n\text{ is even}]\frac{z^n}{n!} \\
&= \sum_{k\ge 0}\left(\frac{2k}{(2k+1)!}\sum_{n>2k} z^n \right) + \frac{e^z + e^{-z}}2 \\
&= \sum_{k\ge 0}\left(\frac{2k}{(2k+1)!}\frac{z^{2k+1}}{1-z} \right) + \frac{e^z + e^{-z}}2 \\
&= \frac{1}{1-z}\sum_{k\ge 0}\left(\frac{2k}{(2k+1)!}z^{2k+1}\right) + \frac{e^z + e^{-z}}2 \\
\end{align}
$$
Now to evaluate the sum above, we start with the known
$$\frac{e^z - e^{-z}}{2} = \sum_{k\ge 0} \frac{z^{2k+1}}{(2k+1)!}$$
and differentiate it:
$$\frac{e^z + e^{-z}}{2} = \sum_{k \ge 0} \frac{(2k+1)z^{2k}}{(2k+1)!}$$
so
$$\begin{align}
\sum_{k \ge 0} \frac{2k}{(2k+1)!} z^{2k+1}
&= \frac{z(e^z + e^{-z})}{2} - \sum_{k \ge 0} \frac{z^{2k+1}}{(2k+1)!} \\
&= \frac{z(e^z + e^{-z})}{2} - \frac{e^{z} - e^{-z}}{2} \\
&= \frac{e^{z}(z-1) + e^{-z}(z+1)}{2}
\end{align}$$
which makes our generating function
$$
\begin{align}
E(z)
&= \frac{1}{1-z}\frac{e^{z}(z-1) + e^{-z}(z+1)}{2} + \frac{e^z + e^{-z}}2 \\
&= \frac{e^{-z}}{2}\left(\frac{z+1}{1-z} + 1\right) \\
&= \frac{e^{-z}}{1-z}
\end{align}
$$
(which unsurprisingly happens to be the same as $D(z)$ the exponential generating function for derangements).
Method 3.
We can use the "symbolic method" from the book Analytic Combinatorics (available online!) by Flajolet and Sedgewick.
Let $\mathcal{H}$ denote the class of "hooks" (or "hockey sticks"), by which I mean permutations that are increasing except for the last element. So it's a set of numbers, arranged in increasing order, followed by some number that is not the maximum. In the notation of the book,
$$\mathcal{H} = \operatorname{S\scriptsize ET}\left(\mathcal{Z}\right) ^\blacksquare\star \mathcal{Z}$$
immediately giving that the EGF of $\mathcal{H}$ is
$$H(z) = \int_0^z (\frac{d}{dt}e^t) t \, dt = e^z(z-1) + 1$$
(You could of course get the same with actual enumeration, using $H_n = (n-1)$ (you only have choice of the last element), but we're trying to avoid explicit counting in this method.)
Let $\mathcal{E}$ be the class of desarrangements, that we want to count. A desarrangement is either a member of $\mathcal{H}$ of odd size (which therefore has its descent in an even position), followed by an arbitrary permutation (and then relabelled), or an increasing permutation (a set of numbers arranged in increasing order) of even size. In notation,
$$\mathcal{E} = \mathcal{H}_{\text{odd}} \star \operatorname{S\scriptsize EQ}(\mathcal{Z}) + \operatorname{S\scriptsize ET}_{\text{even}}(\mathcal{Z})$$
which immediately translates to the EGF
$$\begin{align}
E(z)
&= \frac{H(z) - H(-z)}{2} \frac{1}{1-z} + \frac{e^z + e^{-z}}{2} \\
&= \frac{e^z(z-1) + 1 - e^{-z}(-z-1) - 1}{2}\frac{1}{1-z} + \frac{e^z + e^{-z}}{2}\\
&= \frac{e^{-z}}{1-z}
\end{align}
$$
as before. Even some intermediate expressions turned out the same as in the previous approach, but the difference this time is that we argued "globally" (which correponds to multiplying generating functions etc.) instead of having to count $E_n$ separately for every $n$.
Best Answer
Here’s one way. Start with the recurrence $d_{n+1}=nd_n+nd_{n-1}$, where $d_n$ is the number of derangements of $[n]=\{1,\dots,n\}$. Multiply by $\frac{x^n}{n!}$ and sum over $n\ge 0$:
$$\sum_{n\ge 0}d_{n+1}\frac{x^n}{n!}=\sum_{n\ge 0}nd_n\frac{x^n}{n!}+\sum_{n\ge 0}nd_{n-1}\frac{x^n}{n!}\;.\tag{1}$$
(Here I make the assumption that $d_{-1}=0$: this is consistent with the recurrence, so it causes no problems.) Let $$D(x)=\sum_{n\ge 0}d_n\frac{x^n}{n!}$$ be the exponential generating function in question. Then $$D\,'(x)=\sum_{n\ge 0}nd_n\frac{x^{n-1}}{n!}=\sum_{n\ge 1}d_n\frac{x^{n-1}}{(n-1)!}=\sum_{n\ge 0}d_{n+1}\frac{x^n}{n!}\;,\tag{2}$$
$$xD\,'(x)=x\sum_{n\ge 0}nd_n\frac{x^{n-1}}{n!}=\sum_{n\ge 0}nd_n\frac{x^n}{n!}\;,\tag{3}$$
and $$xD(x)=\sum_{n\ge 0}d_n\frac{x^{n+1}}{n!}=\sum_{n\ge 0}(n+1)d_n\frac{x^{n+1}}{(n+1)!}=\sum_{n\ge 1}nd_{n-1}\frac{x^n}{n!}=\sum_{n\ge 0}nd_{n-1}\frac{x^n}{n!}\;,\tag{4}$$ since by convention $d_{-1}=0$.
Compare $(2),(3)$, and $(4)$ with $(1)$, and you’ll see that
$$D\,'(x)=xD\,'(x)+xD(x)\;,$$ or $$(1-x)D\,'(x)=xD(x)\;.$$ This is a separable differential equation,
$$\frac{D\,'(x)}{D(x)}=\frac{x}{1-x}=-1+\frac1{1-x}\;,$$
which you can now solve for $D(x)$ by first-year calculus.
Here’s another way, quicker but sneakier. For any particular set $K$ of $k$ elements of $[n]$ there are $d_{n-k}$ permutations of $[n]$ that have $K$ as their set of fixed points. There are $\binom{n}k$ such subsets $K$, so there are $\binom{n}kd_{n-k}$ permutations of $[n]$ with exactly $k$ fixed points. Since there are $n!$ permutations of $[n]$ altogether, $$\sum_{k=0}^n\binom{n}kd_{n-k}=n!\;.\tag{5}$$ The lefthand side of $(5)$ is the $n$-th term of the binomial convolution of the sequences $\langle 1,1,1,\dots\rangle$ and $\langle d_n:n\in\Bbb N\rangle$, so the exponential generating function (egf) of the sequence
$$\left\langle\sum_{k=0}^n\binom{n}kd_{n-k}:n\in\Bbb N\right\rangle=\langle n!:n\in\Bbb N\rangle$$
is the product of the egfs of $\langle 1,1,1,\dots\rangle$ and $\langle d_n:n\in\Bbb N\rangle$. Clearly $$\sum_{n\ge 0}n!\frac{x^n}{n!}=\sum_{n\ge 0}x^n=\frac1{1-x}$$ and $$\sum_{n\ge 0}1\cdot\frac{x^n}{n!}=e^x\;,$$ so $$e^xD(x)=\frac1{1-x}\;,$$ and $$D(x)=\frac{e^{-x}}{1-x}\;.$$