The answer is yes if $\Omega$ admits a uniform extension operator for $W^{1,p}$ and $L^p$. I suspect that it is not written down in the Hitchhiker's guide because they seem to avoid interpolation.

Suppose $0 \leq s \leq 1$ and $1 < p < \infty$. We know that $$W^{s,p}(R^d) = \bigl(L^p(R^d),W^{1,p}(R^d)\bigr)_{s,p}$$ (see the tome of Triebel, Ch. 2.8.1, with $W^{s,p} \dot{=} B^s_{p,p}$ there, cf. Remark 4 in Ch. 2.5.1).

Now let $\Omega$ be an extension domain, simultaneously for $W^{1,p}$ and $L^p$ with the same extension operator. Then the above interpolation formula transfers to the function spaces on $\Omega$ via retraction/coretraction (Ch. 1.2.4 in Triebel), that is, $$W^{s,p}(\Omega) = \bigl(L^p(\Omega),W^{1,p}(\Omega)\bigr)_{s,p}$$ for $0 \leq s \leq 1$ (in fact, the extension operator reveals itself to also work for the $W^{s,p}$ scale this way).

Here comes the kicker: Thanks to Rellich-Kondrachov, $W^{1,p}(\Omega)$ embeds compactly into $L^p(\Omega)$ and this is very nicely compatible with interpolation, because it tells us that $$W^{s,p}(\Omega) = \bigl(L^p(\Omega),W^{1,p}(\Omega)\bigr)_{s,p} \hookrightarrow\hookrightarrow W^{s-\epsilon,p}(\Omega) = \bigl(L^p(\Omega),W^{1,p}(\Omega)\bigr)_{s-\epsilon,p}$$ for every $0 < \epsilon \leq s$! (Ch. 1.16.4 in everyone's favorite book). Now you just have to apply the usual embeddings (Ch. 2.8.1) for $W^{s-\epsilon,p}(\Omega)$, so $$W^{s-\epsilon,p}(\Omega) \hookrightarrow W^{s-\delta,q}(\Omega)$$ for $\delta > \epsilon$ and $s - \epsilon - \frac{n}p \geq s-\delta - \frac{n}{q}$, so $q \leq p^*(n,\delta-\epsilon)$. Since $\epsilon$ was arbitrary, you obtain $q < p^*(n,\delta)$.

The answer is yes if $\Omega$ admits a uniform extension operator for $W^{1,p}$ and $L^p$. I suspect that it is not written down in the Hitchhiker's guide because they seem to avoid interpolation.

Suppose $0 \leq s \leq 1$ and $1 < p < \infty$. We know that $$W^{s,p}(R^d) = \bigl(L^p(R^d),W^{1,p}(R^d)\bigr)_{s,p}$$ (see the tome of Triebel, Ch. 2.4.1, with $W^{s,p} \dot{=} B^s_{p,p}$ there, cf. Remark 4 in Ch. 2.5.1).

Now let $\Omega$ be an extension domain, simultaneously for $W^{1,p}$ and $L^p$ with the same extension operator. Then the above interpolation formula transfers to the function spaces on $\Omega$ via retraction/coretraction (Ch. 1.2.4 in Triebel), that is, $$W^{s,p}(\Omega) = \bigl(L^p(\Omega),W^{1,p}(\Omega)\bigr)_{s,p}$$ for $0 \leq s \leq 1$ (in fact, the extension operator reveals itself to also work for the $W^{s,p}$ scale this way).

Here comes the kicker: Thanks to Rellich-Kondrachov, $W^{1,p}(\Omega)$ embeds compactly into $L^p(\Omega)$ and this is very nicely compatible with interpolation, because it tells us that $$W^{s,p}(\Omega) = \bigl(L^p(\Omega),W^{1,p}(\Omega)\bigr)_{s,p} \hookrightarrow\hookrightarrow W^{s-\epsilon,p}(\Omega) = \bigl(L^p(\Omega),W^{1,p}(\Omega)\bigr)_{s-\epsilon,p}$$ for every $0 < \epsilon \leq s$! (Ch. 1.16.4 in everyone's favorite book). Now you just have to apply the usual embeddings (Ch. 2.8.1) for $W^{s-\epsilon,p}(\Omega)$, so $$W^{s-\epsilon,p}(\Omega) \hookrightarrow W^{s-\delta,q}(\Omega)$$ for $\delta > \epsilon$ and $s - \epsilon - \frac{n}p \geq s-\delta - \frac{n}{q}$, so $q \leq p^*(n,\delta-\epsilon)$. Since $\epsilon$ was arbitrary, you obtain $q < p^*(n,\delta)$.