= Solution
The small denominator lies near $x=0$ when $\gamma$ approaches $1$. Its two contributions balance when $\gamma-1=O(\varepsilon)$ and $x=O(\varepsilon)$. Let
$$
\gamma=1+\varepsilon\eta,\qquad x=\varepsilon s,\qquad L=\log2,
$$
with fixed real $\eta$. Then $1-\gamma+\gamma x=\varepsilon(s-\eta)+O(\varepsilon^2)$ and $\log(2-x)=L+O(\varepsilon s)$ in the contributing region. The <endpoint transition of a regularized interior pole> gives
$$
\boxed{I(1+\varepsilon\eta;\varepsilon)\sim\frac1{\varepsilon L}\left[\frac\pi2+\arctan\left(\frac\eta L\right)\right].}
$$
Indeed, the leading scaled <integral> is $\varepsilon^{-1}\int_0^\infty[(s-\eta)^2+L^2]^{-1}ds$. The upper endpoint can be sent to infinity at leading order because the omitted tail contributes only $O(1)$ to the original <integral>. At $\gamma=1$ the result is $\pi/(2\varepsilon\log2)$: only one half of the narrow peak lies in the domain.
This formula provides the required matching. For $\eta\to-\infty$, $\pi/2+\arctan(\eta/L)\sim-L/\eta$, giving $I\sim1/(1-\gamma)$. For $\eta\to+\infty$ it tends to $\pi$, giving $I\sim\pi/(\varepsilon\log2)$, the $\gamma\downarrow1$ limit of the interior-peak approximation. These are <overlap regions> with $\varepsilon\ll|\gamma-1|\ll1$, not assertions that the fixed-$\eta$ remainder is uniform to arbitrarily large $\eta$.
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