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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 $\epsilon\downarrow 0$.
First rewrite the right-hand-side as $$F(x)=c_{1-\epsilon}\int_{-\infty}^\infty \frac{u(x)-u(y)}{|x-y|^{3-\epsilon}}\,dy=\tfrac{1}{2}c_{1-\epsilon}\int_{-\infty}^\infty \frac{2u(x)-u(x+y)-u(x-y)}{|y|^{3-\epsilon}}\,dy,$$ and Fourier transform to obtain $$\hat{F}(k)=c_{1-\epsilon}\int_{-\infty}^\infty e^{ikx}F(x)\,dx=c_{1-\epsilon}\,\hat{u}(k)\int_{-\infty}^\infty \frac{1-\cos ky}{|y|^{3-\epsilon}}\,dy=$$ $$\qquad=c_{1-\epsilon}\,k^2\hat{u}(k)\int_{-\infty}^\infty \frac{1-\cos z}{|z|^{3-\epsilon}}\,dz=c_{1-\epsilon}\,k^2\hat{u}(k) \; 2 \cos( \pi \epsilon/2) \Gamma (\epsilon-2)$$ $$\qquad\rightarrow\frac{c_{1-\epsilon}}{\epsilon}k^2\hat{u}(k) \;\;\text{for}\;\;\epsilon\downarrow 0.$$
So if you let $c_{1-\epsilon}\rightarrow\epsilon$ for $\epsilon\downarrow 0$ you get the desired $\hat{F}(k)\rightarrow k^2\hat{u}(k)$.