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 $$\begin{align*}d_n&=(-1)^nn(n-2)!+(-1)^{n-1}(n-1)!\\
&=(-1)^n\Big(n(n-2)!-(n-1)!\Big)\\
&=(-1)^n(n-2)!\Big(n-(n-1)\Big)\\
&=(-1)^n(n-2)!\;.
\end{align*}$$
You don’t need your expressions for $P_e(x)$ and $P_o(x)$; just expand
$$\prod_{k\ge 1}\frac1{1+x^k}=\prod_{k\ge 1}(1-x^k+x^{2k}-+\ldots\;,$$
and notice that the individual $x^n=x^{k_1}x^{k_2}\ldots x^{k_m}$ terms of the outer product are positive or negative according as $m$ is even or odd, so the coefficient of $x^k$ in
$$\prod_{k\ge 1}\frac1{1+x^k}$$
is the number of partitions of $k$ with an even number of parts minus the number with an odd number of parts.
You may find this question and its answers of interest; it goes beyond your present problem, but I shouldn’t be surprised if you found yourself dealing with the topic soon.
Best Answer
Now, $$ \frac{1}{(1-x)(1+x^2)(1-x^3)\dots} $$
$$ = (1+x+x^2+\dots)(1-x^2+x^4-\dots)(1+x^3+x^6+\dots)(1-x^4+x^8-\dots)\dots $$
Consider a term which provides a negative coefficient to $x^n$
If we pick $x^{2n_2}$ from the second term, $x^{4n_4}$ from the fourth and so on,
The coefficient of $-1$ is $$(-1)^{n_2 + n_4 + \dots}$$
this will be negative if and only if $n_2 + n_4 + \dots$ is odd which comes from $o(n)$.
The positive ones come from $e(n)$.
Thus we must have that
$$\sum (e(n) - o(n))x^n = \frac{1}{(1-x)(1+x^2)(1-x^3)\dots}$$