Here is a derivation by Fourier transformation of the right-hand-side for $s=1-\epsilon$ to check that it equals $k^2 \hat{u}(k)$ when $\epsilon\downarrow 0$.
I will make use of the fact that the coefficient $c_s$, given in Wikipedia, has the expansion
$$c_{1-\epsilon}=2\epsilon+{\cal O}(\epsilon^2).$$
First rewrite the right-hand-side as $$F_\epsilon(x)\equiv c_{1-\epsilon}\int_{-\infty}^\infty \frac{u(x)-u(y)}{|x-y|^{3-2\epsilon}}\,dy=\tfrac{1}{2}c_{1-\epsilon}\int_{-\infty}^\infty \frac{2u(x)-u(x+y)-u(x-y)}{|y|^{3-2\epsilon}}\,dy,$$ and Fourier transform to obtain $$\hat{F}_\epsilon(k)\equiv\int_{-\infty}^\infty e^{ikx}F_\epsilon(x)\,dx=c_{1-\epsilon}\,\hat{u}(k)\int_{-\infty}^\infty \frac{1-\cos ky}{|y|^{3-2\epsilon}}\,dy=$$ $$\qquad=c_{1-\epsilon}\,k^2\hat{u}(k)\int_{-\infty}^\infty \frac{1-\cos z}{|z|^{3-2\epsilon}}\,dz=c_{1-\epsilon}\,k^2\hat{u}(k) \; 2 \cos( \pi \epsilon) \Gamma (2\epsilon-2)$$ $$\qquad\rightarrow k^2\hat{u}(k) \;\;\text{for}\;\;\epsilon\downarrow 0,$$ since $ 2 \cos( \pi \epsilon) \Gamma (2\epsilon-2)=\frac{1}{2\epsilon}+{\cal O}(1)$.