Well, technically it's those functions $z \in C^\infty(\mathbb{R}^{n+1}_+)$ whose support are contained inside the region bounded by $\partial_L C_\Omega$, to use your notation.
In that case, the answer is yes, and you do the mollification in the usual way (the following is a straightforward recounting of the mollification technique that you can find, for example, in the first chapter of Adams' book on Sobolev Spaces, slightly adapted for our case).
Let $u \in X_0^\alpha(C_\Omega)$, and $\varphi$ be your favorite bump function supported on a ball of radius 1 in $\mathbb{R}^{n+1}$, say, and make your space the whole cylinder $C_\Omega' = \{ (x,y): x \in \Omega, y \in \mathbb{R} \}$ and extend $u$ by even reflection (for $y < 0$, let $u(x,y) = u(x,-y)$). Let $d(X) = d(x,\partial \Omega)$ where $X=(x,y)$. Define $u(X,\epsilon) = \begin{cases}u(X) & \text{if} \,\,d>2\epsilon \\ 0 &\text{otherwise}\end{cases} $ and then subsequently $u_\epsilon(X) = (u(X,\epsilon) *\frac{1}{\epsilon^n}\varphi(\frac{X}{\epsilon}))$ From here, it's fairly simple to see that $u_\epsilon \in C^\infty_c(C_\Omega')$, and it remains only to show that $\int_{C_\Omega'} y^{1-\alpha} |\nabla u - \nabla u_\epsilon|^2 dx dy \rightarrow 0$ as $\epsilon \rightarrow 0$.
Notice that $y^{1-\alpha} dx dy$ is a nice bounded positive measure for $0 < \alpha < 2$ (which is always the case when you consider the extension for the fractional Laplacian), hence continuous functions of compact support are dense in $L^2(y^{1-\alpha} dx dy)$. Choose $n+1$ such nice functions $G=(g_1,\ldots,g_{n+1})$, such that $\int_{C_\Omega'} |\nabla u - G|^2 y^{1-\alpha} dx dy < \frac{\eta}{6}$ Define $G_\epsilon(X)$ the same way we defined $u_\epsilon$. Now $|\nabla u - \nabla u_\epsilon| \leq |\nabla u - G| + |G - G_\epsilon| + |G_\epsilon - \nabla u_\epsilon|$ It is clear (by simply using Fubini and doing the convolution first) that $\int y^{1-\alpha} |G_\epsilon - \nabla u_\epsilon|^2 dx dy < \frac{\eta}{6}$ so we are left only to consider the term in $|G_\epsilon - G|$. However, $G$ is uniformly continuous, hence for $\epsilon$ sufficiently small we have $|G_\epsilon - G|^2 < \frac{\eta}{C}$, except possibly when $d(X) < 3\epsilon$. Since the support of $G$ is bounded, pick $C$ so that $\int y^{1-\alpha} |G_\epsilon - G|^2 dx dy < \frac{\eta}{6}$