If $C$ is any set of positive integers, and $g_C(n)$ is the number of permutations of $[n]$ whose cycle lengths are all in $C$, then $$G_C(x)=\sum_{n\ge 0}g_C(n)\frac{x^n}{n!}=\exp\left(\sum_{n\in C}\frac{x^n}{n}\right)$$ is the exponential generating function for the $g_C(n)$. (Rather than derive it, I’ve simply quoted this from Theorem 4.34 in Miklós Bóna, Introduction to Enumerative Combinatorics.)
In your case $C=\mathbb{Z}^+\setminus\{2\}$, so it’s $$\begin{align*}
G_C(x)&=\exp\left(\sum_{n\ge 1}\frac{x^n}n-\frac{x^2}2\right)\\
&=\exp\left(-\ln(1-x)-\frac{x^2}2\right)\\
&=\frac{e^{-x^2/2}}{1-x},
\end{align*}$$ and your generating function is correct. Then
$$\frac{e^{-x^2/2}}{1-x}=\left(\sum_{n\ge 0}x^n\right)\left(\sum_{n\ge 0}\frac{(-1)^nx^{2n}}{n!2^n}\right),$$ and
$$[x^n]\left(\sum_{n\ge 0}x^n\right)\left(\sum_{n\ge 0}\frac{(-1)^nx^{2n}}{n!2^n}\right)=\sum_{k=0}^{\lfloor n/2\rfloor}\frac{(-1)^k}{k!2^k}.$$
Recall, though, that the coefficient of $x^n$ in $G_C(x)$ is not $g_C(n)$, but rather $\dfrac{g_C(n)}{n!}$, so $$g_C(n)=n!\sum_{k=0}^{\lfloor n/2\rfloor}\frac{(-1)^k}{k!2^k}.$$
As a quick check, this yields $g_C(1)=1$, $g_C(2)=2\left(1-\frac12\right)=1$, $g_C(3)=6\left(1-\frac12\right)=3$, $g_C(4)=24\left(1-\frac12+\frac18\right)=15$, and $g_C(5)=120\left(1-\frac12+\frac18\right)=75$, all of which are in agreement with the OEIS values.
For your inclusion-exclusion argument, $c_k=0$ for $k>\lfloor n/2\rfloor$, and $c_0=n!$, so what you actually want is $$\sum_{k=0}^{\lfloor n/2\rfloor}(-1)^kc_k=\sum_{k=0}^{\lfloor n/2\rfloor}(-1)^k\frac{n!}{k!2^k},$$ which is exactly what we just got with generating functions.
Here is an argument completely different from joriki’s; it is also a complete solution.
A cycle is an odd permutation iff its length is even, so a permutation written as a product of disjoint cycles is even iff an even number of the factors are even cycles. Let $\pi=\sigma_1\dots\sigma_k$ be a permutation of $[n]$ written as a product of $k$ disjoint cycles. Then $n=|\,\sigma_1|+\dots+|\,\sigma_k|$, so the parity of $n$ is the same as the parity of the number of cycles of odd length. Thus, $\pi$ has an even number of cycles of even length iff $n$ and $k$ have the same parity, i.e., iff $(-1)^{n+k}=1$.
Let $e_n$ be the total number of cycles in even permutations of $[n]$, let $o_n$ be the total number of cycles in odd permutations of $[n]$, and let $d_n=e_n-o_n$. The total number of permutations of $[n]$ with $k$ cycles is given by $\left[n\atop k\right]$, the unsigned Stirling number of the first kind. Each of these $\left[n\atop k\right]$ permutations contributes $k$ cycles to $e_n$ if $(-1)^{n+k}=1$, and to $o_n$ if $(-1)^{n+k}=(-1)$. Thus, each contributes $(-1)^{n-k}k$ to $d_n$, and it follows that
$$d_n=\sum_k(-1)^{n+k}k\left[n\atop k\right]\;.$$
Now $(-1)^{n+k}\left[n\atop k\right]=(-1)^{n-k}\left[n\atop k\right]$ is the signed Stirling number of the first kind, for which we have the generating function $$\sum_k(-1)^{n+k}\left[n\atop k\right]x^k=x^{\underline{n}}\;.\tag{1}$$
(Here $x^{\underline{n}}=x(x-1)(x-2)\cdots(x-n+1)$ is the falling factorial, sometimes written $(x)_n$.)
Differentiate $(1)$ with respect to $x$ to obtain
$$\sum_k(-1)^{n+k}k\left[n\atop k\right]x^{k-1}=Dx^{\underline{n}}=(x-n+1)Dx^{\underline{n-1}}+x^{\underline{n-1}}\;,$$
where the last step is simply the product rule, since $x^{\underline{n}}=x^{\underline{n-1}}(x-n+1)$. But $$Dx^{\underline{n-1}}=\sum_k(-1)^{n-1+k}k\left[{n-1}\atop k\right]x^{k-1}\;,$$ so
$$\sum_k(-1)^{n+k}k\left[n\atop k\right]x^{k-1}=\sum_k(-1)^{n-1+k}k\left[{n-1}\atop k\right]x^{k-1}+x^{\underline{n-1}}\;.\tag{2}$$
Now evaluate $(2)$ at $x=1$ to get $$d_n=(2-n)d_{n-1}+[n=1]\;,\tag{3}$$ where the last term is an Iverson bracket. If we set $d_0=0$, $(3)$ yields $d_1=[1=1]=1$, which by direct calculation is the correct value: the only permutation of $[1]$ is the identity, which is even and has one cycle. It’s now a trivial induction to check that $d_n=(-1)^n(n-2)!$ for all $n\ge 1$: the induction step is
$$\begin{align*}d_{n+1}&=\Big(2-(n+1)\Big)d_n\\
&=(1-n)(-1)^n(n-2)!\\
&=(-1)^{n+1}(n-1)!\;.
\end{align*}$$
Added: It occurs to me that there’s a rather easy argument that does not use generating functions. Let $\sigma$ be any $k$-cycle formed from elements of $[n]$; then $\sigma$ is a factor in $(n-k)!$ permutations of $[n]$. Moreover, exactly half of these permutations are even unless $n-k$ is $0$ or $1$. Thus, $\sigma$ contributes to $d_n$ iff $k=n$ or $k=n-1$.
There are $(n-1)!$ $n$-cycles; they are even permutations iff $n$ is odd, so they contribute $(-1)^{n-1}(n-1)!$ to $d_n$.
There are $n(n-2)!$ $(n-1)$-cycles: there are $n$ ways to choose the element of $n$ that is not part of the $(n-1)$-cycle, and the other $n-1$ elements can be arranged in $(n-2)!$ distinct $(n-1)$-cycles. The resulting permutation of $[n]$ is even iff $n-1$ is odd, i.e., iff $n$ is even, so they contribute $(-1)^nn(n-2)!$ to $d_n$.
It follows that $$\begin{align*}d_n&=(-1)^nn(n-2)!+(-1)^{n-1}(n-1)!\\
&=(-1)^n\Big(n(n-2)!-(n-1)!\Big)\\
&=(-1)^n(n-2)!\Big(n-(n-1)\Big)\\
&=(-1)^n(n-2)!\;.
\end{align*}$$
Best Answer
There are two possible interpretations here, the first, permutations consisting of an even number of odd cycles and some even cycles and second, permutations consisting of an even number of odd cycles only.
First interpretation. Observe that the generating function of permutations with odd cycles marked is $$G(z, u) = \exp\left(\sum_{k\ge 1} \frac{z^{2k}}{2k} + u \sum_{k\ge 0}\frac{z^{2k+1}}{2k+1}\right).$$ This is $$G(z, u) = \exp\left((1-u)\sum_{k\ge 1} \frac{z^{2k}}{2k} + u \sum_{k\ge 1}\frac{z^{k}}{k}\right).$$
To get the permutations with an even number of odd cycles use $$\frac{1}{2} G(z,1)+\frac{1}{2} G(z, -1)$$ which yields $$\frac{1}{2}\exp\left(\sum_{k\ge 1}\frac{z^{k}}{k}\right) + \frac{1}{2} \exp\left(2\sum_{k\ge 1} \frac{z^{2k}}{2k} - \sum_{k\ge 1}\frac{z^{k}}{k}\right).$$
This simplifies to $$\frac{1}{2} \frac{1}{1-z} + \frac{1}{2} (1-z) \frac{1}{1-z^2}$$ which is $$\frac{1}{2} \frac{1}{1-z} + \frac{1}{2} \frac{1}{1+z}.$$
This simply says that when $n$ is even then there must be an even number of odd cycles and when $n$ is odd there cannot be an even number of odd cycles, which follows by inspection (parity).
Second interpretation. Here we have $$G(z, u) = \exp\left(u \sum_{k\ge 0}\frac{z^{2k+1}}{2k+1}\right).$$ This is $$G(z, u) = \exp\left(u \sum_{k\ge 0}\frac{z^{k}}{k} - u \sum_{k\ge 1}\frac{z^{2k}}{2k}\right).$$ To get the permutations with an even number of odd cycles use $$\frac{1}{2} G(z,1)+\frac{1}{2} G(z, -1)$$ which yields $$\frac{1}{2}\frac{1}{1-z} \left(\frac{1}{1-z^2}\right)^{-1/2} + \frac{1}{2} (1-z) \left(\frac{1}{1-z^2}\right)^{1/2}.$$
This gives the sequence $$0, 1, 0, 9, 0, 225, 0, 11025, 0, 893025, 0, 108056025, 0, 18261468225,\ldots$$ which points us to OEIS A177145, where we find this computation confirmed.
This generating function maybe written as $$\frac{1}{2}\frac{1}{1-z} \sqrt{1-z^2} + \frac{1}{2} (1-z) \frac{1}{\sqrt{1-z^2}}$$ or $$\frac{1}{2}\sqrt{\frac{1+z}{1-z}} + \frac{1}{2}\sqrt{\frac{1-z}{1+z}}.$$