What you are doing is integrating ${1 \over |1 - re^{i\theta}|}$ over the unit circle, where $r = e^{-2\pi y}$. Geometrically $|1 - re^{i\theta}|$ is the distance between $z = re^{i\theta}$ and $z = 1$. When $|\theta| < 1 - r $, this distance is $O(1 - r)$, so the contribution to the integral for $|\theta| < 1 - r$ will be $C{1 \over 1 - r}* 1 - r < C$.
When $|\theta| > 1 - r$, the distance is of the same order of magnitude as the distance to $z = r$ to $z = r^{i\theta}$. Since $r$ is near 1, your integral is the same order of magnitude as $ \int_{|\theta| > 1 - r}{1 \over |1 - e^{i\theta}|}$ $= \int_{|\theta| > 1 - r}\frac{1}{|e^{-i{\theta \over 2}} - e^{i{\theta \over 2}}|}$ $= \int_{|\theta| > 1 - r}\frac{1}{2|\sin{\theta \over 2}|}$ Since $\sin(\theta) \sim \theta$, you are integrating ${1 \over \theta}$ basically and your term becomes $O(\ln (1 - r))$. So the overall integral is $O(\ln (1 - r))$. Plugging back in $r = e^{-2\pi y}$ this is $O(\ln (1 - e^{-2\pi y}))$ or $O (\ln(y))$ if you plug in the Taylor expansion.
If you want to go from this to $<< |\ln(y)|$ instead of $O(\ln(y))$, you can use the fact the first term is a lot smaller than the second. So instead of breaking up at $|\theta| = 1 - r$, break it up at $|\theta| = N(1 - r)$ for some large $N$. You will get a correspondingly smaller constant in front of $|\ln(y)|$ in the second term, while the first term is still much smaller as $y$ gets small.