$\newcommand\ep\epsilon\newcommand\Om\Omega\newcommand\al\alpha\newcommand\R{\mathbb R}$Your desired conclusion is true. Indeed, take any $u\in C^{0,s}(\R^n)$ such that $u$ is Hölder-continuous on $\Om$ with exponent $2s+\ep\in(0,1)$. Then $u$ is continuous on $\R^n$ (which is all we need in place of the condition $u\in C^{0,s}(\R^n)$). 

It follows that $u\in C^{0,2s+\ep}(\R^n)$. Indeed, we know that $u$ is Hölder-continuous on $\Om$ with exponent $2s+\ep\in(0,1)$, that is, 
$$|u(x)-u(y)|\le c|x-y|^{2s+\ep}\tag{1}$$
for some real $c>0$ and all $x,y$ in $\Om$. Being continuous, $u$ is also Hölder-continuous on the closure of $\Om$ with exponent $2s+\ep\in(0,1)$. 
Also, $u$ is Hölder-continuous on $\R^n\setminus\Om$ with any exponent, because $u=0$ on $\R^n\setminus\Om$. To show that $u\in C^{0,2s+\ep}(\R^n)$, it remains to show that (1) holds for any $x\in\Om$ and $y\in\R^n\setminus\Om$. Take any such $x,y$. On the straight line segment connecting $x$ and $y$, there is a point $z$ lying on the boundary of $\Om$. Then $|x-z|\le|x-y|$ and, by the continuity of $u$, we have $u(z)=0$, so that $u(z)=u(y)$ and hence, by (1) 
$$|u(x)-u(y)|=|u(x)-u(z)|\le c|x-z|^{2s+\ep}\le c|x-y|^{2s+\ep}.$$
This verifies that $u\in C^{0,2s+\ep}(\R^n)$. 

Now your desired conclusion follows by the first, "positive" part of the [previous answer][1]. 

  [1]: https://mathoverflow.net/a/376532/36721