A sketch of a proof that $c(n)=p_{DO}(n)$ can be found here; it involves a simple manipulation of Ferrers diagrams. The rest seems easiest to do with generating functions.
First note that in
$\prod_{k\ge 1}\frac1{1+x^k}=\prod_{k\ge 1}(1-x^k+x^{2k}-+\dots)\;,$
the individual $x^n=x^{k_1}x^{k_2}\dots x^{k_m}$ terms are positive or negative according as $m$ is even or odd, and therefore
$\prod_{k\ge 1}\frac1{1+x^k}=\sum_{k\ge 0}\big(p_E(k)-p_O(k)\big)x^k\,:$
a partition $(k_1,k_2,\dots,k_m)$ is counted positively when $m$ is even and negatively when $m$ is odd.
Similarly, in
$\prod_{k\ge 0}(1-x^{2k+1})$
the $x^n$ terms are all of the form $x^{2k_1+1}x^{2k_2+1}\dots x^{2k_m+1}$, where the $k_i$ are distinct, and the term is positive for even $m$ and negative for odd $m$, so
$\prod_{k\ge 0}(1-x^{2k+1})=\sum_{k\ge 0}(-1)^kp_{DO}(k)x^k\;.$
If we can show that
$\prod_{k\ge 0}(1-x^{2k+1})=\prod_{k\ge 1}\frac1{1+x^k}\;,\tag{1}$
we’ll be able to conclude that $p_E(k)-p_O(k)=(-1)^kp_{DO}(k)$ and hence that $p_{DO}(k)=(-1)^k(p_E(k)-p_O(k))$.
To prove $(1)$, observe that
$\begin{align*} \prod_{k\ge 1}\frac1{1+x^k}&=\prod_{k\ge 1}\frac{1-x^k}{1-x^{2k}}\\\\ &=\frac{\prod\limits_{k\ge 1}(1-x^k)}{\prod\limits_{k\ge 1}(1-x^{2k})}\\\\ &=\frac{\prod\limits_{k\ge 1}(1-x^{2k})\prod\limits_{k\ge 0}(1-x^{2k+1})}{\prod\limits_{k\ge 1}(1-x^{2k})}\\\\ &=\prod_{k\ge 0}(1-x^{2k+1})\;. \end{align*}$