Yes, this follows from the identity
$$(-\Delta^{s}u)(x)=c_{n,s}\int_{\mathbb{R}^n}\frac{u(x)-u(y)}{||x-y||^{n+2s}}\,dy,$$
$$c_{n,s}=\frac{s\,4^{s}\Gamma(s+n/2)}{\pi^{n/2}\Gamma(1-s)},$$
where the principal value of the integral should be taken. (See The Pohozaev identity for the fractional Laplacian.)
Integration over $x\in\mathbb{R}^n$ gives zero because of the antisymmetry of the integrand upon interchange $x\leftrightarrow y$, so
$$\int_{\mathbb{R}^n} [(-\Delta^{s}u)(x)+u(x)]\,dx=\int_{\mathbb{R}^n} u(x)\,dx=\int_{\mathbb{R}^n} f(u(x))\,dx,$$
as requested.