We prove by induction that $o(n) = 2^{2n-1} - 2^{n-1}$. For $n = 1$, this reads $o(1) = 2^1 - 2^0 = 1$, which is obviously true - the only such 2-tuple is $(1,1)$. Suppose it is true for $n$.

If $(a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n)$ gives an odd sum, then we can take $(a_{n+1}, b_{n+1})$ to be any of $(0, 0), (0, 1), (1, 0)$ and still get an odd sum for $(a_1, a_2, \ldots, a_{n+1}, b_1, b_2, \ldots, b_{n+1})$. On the other hand, if $(a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n)$ is even, then we must have $a_{n+1} = b_{n+1} = 1$ to get an odd sum. Thus $o(n+1) = 3o(n) + e(n)$. But $o(n) = 2^{2n-1} - 2^{n-1}$ and $e(n) = (o(n) + e(n)) - o(n) = 2^{2n} - 2^{2n-1} + 2^{n-1} = 2^{2n-1} + 2^{n-1}$. So $o(n+1) = 4 \cdot 2^{2n-1} - 2 \cdot 2^{n-1} = 2^{2(n+1)-1} - 2^{(n+1)-1}$, which establishes the result for $n+1$ and hence for all $n$.

Hence $e(n) = 2^{2n} - o(n) = 2^{2n-1} + 2^{n-1}$ and $\frac{o(n)}{e(n)} = \frac{2^n - 1}{2^n + 1}$.