Using Kahan summation algorithm should suffice your purpose. Having:
$$
R=\frac{\sum\limits_{i=1}^{N}a_i}{\sum\limits_{i=1}^{N}b_i}=\frac{A}{B}
$$
if $A$ and $B$ are computed accurate enough, the final ratio $R$ should not be a problem.
If $A$ and $B$ are computed using Kahan summation, you can expect the relative error to be $\mathcal{O}(\epsilon_\text{mach}+N\epsilon_\text{mach}^2)$. For double-precision and $N\approx 10^{10}$ you are safe ($N\ll 1/\epsilon_\text{mach}$), and your error is practically independent of $N$: $\mathcal{O}(\epsilon_\text{mach})$.
It's worth to mention, that the constant in front of the error estimate (condition # of summation $\kappa_{\sum}$) can still be relatively large.
If summation condition number:
$$
\kappa_{\sum_A}=\frac{\sum\limits_{i=1}^{N}|a_i|}{\left|\sum\limits_{i=1}^{N}a_i\right|}
$$
is large, your best bets would probably be in Shewchuk's algorithm. With that, you will be able to calculate your numerator and denominator to full double precision.
If $\tilde{A}$ and $\tilde{B}$ are calculated in full double precision (say using Shewchuk's algorithm):
$$
\tilde{A}=A(1+\delta_A),\quad \tilde{B}=B(1+\delta_B),\quad |\delta_A|,|\delta_B|\le\epsilon_\text{mach}
$$
, then
$$
\begin{aligned}
\tilde{R}&=\tilde{A}\oslash\tilde{B}&=\frac{A(1+\delta_A)}{B(1+\delta_B)}(1+\delta_{\oslash})=\frac{A}{B}(1+\theta)\\
&&|\theta|\leq\frac{3\epsilon_\text{mach}}{1-3\epsilon_\text{mach}}
\end{aligned}
$$
Here, $\tilde{A},\tilde{B},\tilde{R}$ are floating point (IEEE-754) representation of $A$, $B$, and $R$, respectively, $\oslash$ - floating point analog of division.
Now, even if $A$ and $B$ are not calculated to the full precision (Kahan), it will just increase $\delta_A$ and $\delta_B$ (not allowing approximation by $\theta$ with such a tight bound). However, I still do not see how the division can be the problem here. Especially, since the true expected values of $A$ and $B$ are in $[10^{-1}, 10^{1}]$.
