For the first equality, take the Young diagram of a self-conjugate partition and break it into hooks with a diagonal square as corner; their sizes form a partition of $n$ into distinct odd parts, and the inverse operation is obvious.
For the second equality, consider the following involution to match any partition of $n$ in the subset $S=p(n)\setminus p_{DO}(n)$ of partitions that are not into distinct odd parts, with another element of $S$ with the opposite parity of its number of its parts. For $\lambda\in S$, test the odd numbers $m=1,3,5,\ldots$ in order, considering the set of parts of $\lambda$ of the form $2^km$ with $k\in\mathbf N$. If the set is either empty or consists of a single part $m$, move on to the next odd number. Since $\lambda\in S$, there is some $m$ that is not skipped; stop at the smallest such $m$. Now take the maximal $k$ such that $\lambda$ has a part of size $2^km$. If this part is unique, then $k\neq0$ by the choice of $m$; break the part into two parts of size $2^{k-1}m$. Otherwise $\lambda$ has at least two such parts; glue them together to form a part of size $2^{k+1}m$. One readily checks that the partition $\mu$ so produced is in $S$, that this procedure is an involution on $S$, and that $\lambda,\mu$ always have opposite parities for the number of their parts.
So the elements of $S$ together produce a null contribution to $p_E(n)-p_O(n)$. It remains to show that each element of $p_{DO}(n)$ contributes $(-1)^n$ to $p_E(n)-p_O(n)$. But it is obvious (by adding up parts modulo $2$) that the parity of the number of parts of any partition in $P_{DO}(n)$ is the same as that of $n$, which settles this point.
Best Answer
For the sake of completeness here is a slightly different approach to part one, which then continues along the path in the link from Brian Scott.
The generating function of the set of partitions by the number of parts is given by $$ G(z,u) = \prod_{k\ge 1} \frac{1}{1-uz^k}.$$ It follows that the generating function of the set of partitions with an even number of parts is given by $$ G_1(z) = \frac{1}{2} G(z, 1) + \frac{1}{2} G(z, -1) = \frac{1}{2} \prod_{k\ge 1} \frac{1}{1-z^k} + \frac{1}{2} \prod_{k\ge 1} \frac{1}{1+z^k}.$$ Similarly for an odd number of parts, $$ G_2(z) = \frac{1}{2} G(z, 1) - \frac{1}{2} G(z, -1) = \frac{1}{2} \prod_{k\ge 1} \frac{1}{1-z^k} - \frac{1}{2} \prod_{k\ge 1} \frac{1}{1+z^k}.$$ Therefore $$ G_1(z) - G_2(z) = Q(z) = \prod_{k\ge 1} \frac{1}{1+z^k}$$ and this is the generating function of the difference between the number of partitions into an even and odd number of parts.
Now observe that this is $$\prod_{k\ge 1} \frac{1-z^k}{1-z^{2k}} = \prod_{k\ge 0} (1-z^{2k+1})$$ because the denominator cancels all the factors with even powers in the numerator.
Here is an important observation: the above expression of $Q(z)$ enumerates partitions into unique odd parts in a generating function of signed coefficients where the sign indicates the parity of the number of parts. There is no cancellation between partitions that add up to the same value because the counts of constituent parts have the same parity. Moreover as all parts are odd, partitions of odd numbers must have an odd number of parts and of even numbers, an even number. Therefore the coefficients of $Q(z)$ alternate in sign.
To get the series that generates the absolute values of these coefficients, we create generating functions for the even powers and the odd ones, inverting the sign of the coefficients of the odd ones. The even ones are generated by $$\frac{1}{2} Q(z) + \frac{1}{2} Q(-z)$$ and the odd ones by $$-\left(\frac{1}{2} Q(z) - \frac{1}{2} Q(-z)\right)$$ Adding these gives $$ Q(-z).$$ But this is $$\prod_{k\ge 0} (1-(-z)^{2k+1}) = \prod_{k\ge 0} (1-(-1)^{2k+1} z^{2k+1}) = \prod_{k\ge 0} (1+ z^{2k+1}),$$ precisely the generating function of partitions into distinct odd parts.
This is sequence A000700 from the OEIS.