This is a very nice question. It appears to me that the answer is in general, no. Moreover, it appears to be an open problem determining geometric conditions on $\Omega$ for which such an extension holds. Essentially, this amounts to establishing fractional boundary Hardy inequality. Indeed, take $u\in C^\infty_0(\Omega)$ and let $$Eu(x) = \begin{cases}
u(x), &\text{if } x\in \Omega,\\
0, &\text{else}
\end{cases}
$$ i.e. extend by zero. Let $0<s<1$ and $1\leq p <+\infty$. Then $\| Eu\|_{L^p(\mathbb R^n)}=\| u\|_{L^p(\Omega)}$ and \begin{align*}
[ Eu ]_{W^{s,p}(\mathbb R^n)}^p &= \int_{\mathbb R^n}\int_{\mathbb R^n} \frac{\vert Eu(x)-Eu(y)\vert^p}{\vert x- y \vert^{n+sp}}\, dy \, dx \\
&=\int_{\Omega}\int_{\Omega} \frac{\vert u(x)-u(y)\vert^p}{\vert x- y \vert^{n+sp}}\, dy \, dx + 2\int_{\mathbb R^n\setminus \Omega}\int_{\Omega} \frac{\vert u(y)\vert^p}{\vert x- y \vert^{n+sp}}\, dy \, dx\\
&= [ u ]_{W^{s,p}(\Omega)}^p+ 2\int_{\mathbb R^n\setminus \Omega}\int_{\Omega} \frac{\vert u(y)\vert^p}{\vert x- y \vert^{n+sp}}\, dy \, dx\\
&=[ u ]_{W^{s,p}(\Omega)}^p+ 2\int_{\Omega} \vert u(y)\vert^p w(y)\, dy
\end{align*} where $$ w(y) = \int_{\mathbb R^n\setminus \Omega}\frac{ dx}{\vert x- y \vert^{n+sp}}.$$
To prove $ \|Eu\|_{W^{s,p}(\mathbb R^n)} \leq C\|u\|_{W^{s,p}(\Omega)} $ with $C>0$ depending only on $n$, $s$, $p$ (what we need to show to conclude $E$ is bounded), we must show that $$\int_{\Omega} \vert u(y)\vert^p w(y)\, dy \leq C \|u\|_{W^{s,p}(\Omega)} $$ for all $u\in C^\infty_0(\Omega)$. For $\Omega$ open, we have that $w(y) \leq C \operatorname{dist}(y,\mathbb R^n \setminus \Omega)^{-sp}$ (I have provided a proof of this fact below), so if one can show that $$\int_{\Omega} \frac{\vert u(y)\vert^p}{\operatorname{dist}(y,\mathbb R^n \setminus \Omega)^{sp}}\, dy \leq C \|u\|_{W^{s,p}(\Omega)} \tag{$\ast$}$$ for all $u\in C^\infty_0(\Omega)$ then we are done. This is the fractional boundary Hardy inequality which, as I mentioned before, is an open inequality in general.
Moreover, if $\Omega=B_1$ then I expect that we also have that $w(y) \geq C \operatorname{dist}(y,\mathbb R^n \setminus \Omega)^{-sp}$, so proving $E$ is bounded becomes equivalent to proving ($\ast$). However, it was proven by Dyda in A fractional order Hardy inequality that for a bounded Lipschitz domain (which $B_1$ is) the fractional boundary Hardy inequality holds if and only if $sp>1$. To read more, you could look at Fractional Hardy inequalities and visibility of the boundary by Ihnatsyeva, Lehrbäck, Tuominen, and Vähäkangas.
By the Layer-cake theorem, \begin{align*}
w(y)&=(n+sp)\int_0^{+\infty} t^{n+sp-1}\vert \{x\in \mathbb R^n\setminus \Omega \text{ s.t. } \vert x-y\vert^{-1} >t\} \vert \, dt \\
&= (n+sp)\int_0^{+\infty}t^{n+sp-1} \vert ( \mathbb R^n\setminus \Omega ) \cap B_{t^{-1}}(y) \vert \, dt .
\end{align*} When $t>\operatorname{dist}(y,\mathbb R^n \setminus \Omega)^{-1}$ then $( \mathbb R^n\setminus \Omega ) \cap B_{t^{-1}}(y) =\varnothing$, so actually \begin{align*}
w(y)&= (n+sp)\int_0^{\operatorname{dist}(y,\mathbb R^n \setminus \Omega)^{-1}} t^{n+sp-1}\vert ( \mathbb R^n\setminus \Omega ) \cap B_{t^{-1}}(y) \vert \, dt .
\end{align*} Hence, $\vert ( \mathbb R^n\setminus \Omega ) \cap B_{t^{-1}}(y) \vert \leq Ct^{-n}$, so $$ w(y)\leq C\int_0^{\operatorname{dist}(y,\mathbb R^n \setminus \Omega)^{-1}} t^{sp-1} \, dt = C \operatorname{dist}(y,\mathbb R^n \setminus \Omega)^{-sp}.$$
