Let $D(n) = E(n) - O(n)$. We trivially have $O(1) = 1$ and $E(1) = 0$, thus $D(1) = -1$, and $E(2) = O(2) = 1$ (respectively $2 = 1+1$ and $2 = 2$), hence $D(2) = 0$. We will show by total induction that $D(n) = 0$ for all $n > 2$. Assume it holds for all numbers from $2$ to $n-1$. If $n$ is odd, a partition must contain at least one $1$. Since the addition of $1$ changes the parity of the number of one's it follows that $E(n) = O(n-1)$ and $O(n) = E(n-1)$, hence since $D(n) = -D(n-1) = 0$. If $n$ is even the partition must contain an even number $2k$ of $1$s. If it contains $2k = n$ or $2k = n-2$, then we have the unique solutions
$$1+1+\cdots+1 \text{ and } 2+1+1+\cdots+1,$$
the first adding to $E(n)$, the second to $O(n)$. For other, smaller, values of $k$ we note that the remaining exponents are all even, and we can thus apply the inductive hypothesis to $\frac{n-2k}{2}$. Thus, it follows that for even $n$ we will also have $D(n) = 0$. This completes the proof. $\square$

Let $G(x) = 1 + \sum_{i=1}^{\infty} D(i)x^i$ be the generating function for the sequence we desire. To count partitions with an odd number of odd-appearing exponents as negative, we will multiply those choices as negative. Thus, we have that
$$G(x) = (1-x+x^2-x^3+x^4-\dots)(1-x^2+x^4-x^6+x^8-\dots)(1-x^4+x^8-x^{12}+x^{16}-\dots)\dots,$$
where the $i$-term corresponds to choosing the number of times $2^{i-1}$ is represented in the partition. Taking the expression for each power series, one now finds
$$G(x) = \frac{1}{1+x} \frac{1}{1+x^2} \frac{1}{1+x^4} \frac{1}{1+x^8} \dots$$
However,
$$\begin{aligned} \frac{G(x)}{1-x} &= \frac{1}{1-x} \frac{1}{1+x} \frac{1}{1+x^2} \cdots = \frac{1}{1-x^2} \frac{1}{1+x^2} \frac{1}{1+x^4} \cdots \\ &= \frac{1}{1-x^4} \frac{1}{1+x^4} \frac{1}{1+x^8} \cdots = \cdots = 1. \end{aligned}$$
Thus, $G(x) = 1-x$ from which it follows that $D(1) = -1$ and $D(n) = 0$ for all $n > 1$. $\square$