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 $\hat{F}(k)\rightarrow k^2\hat{u}(k)$.