Here's a proof assuming basic properties of the cotangent complexes of stack. At least for the proof to work I need to assume that $X_0$ is smooth over $k$ and the source/target maps $s, t: X_1 \to X_0$ of the groupoid structure are smooth (implying that $X_1$ is also smooth over $k$).

Note: when I say "concentrated in degree 0" I mean "has cohomology only in degree 0."

There's a cartesian diagram of the form $$ \begin{equation} \begin{matrix} X_1 & \xrightarrow{t} & X_0 \\ s \downarrow & & \downarrow \pi \\ X_0 & \xrightarrow{\pi} \, \, \, \mathcal{X} \\ \end{matrix} \end{equation} $$

By the construction of $\mathcal{X}$, $\pi$ is a smooth cover. It follows that $$ \begin{equation} t^* L_{\pi} = L_{s} \end{equation} $$ Since $s$ is smooth by hypothesis, $L_s \simeq \Omega_{X_1/X}[0]$. So in particular $L_s$ is concentrated in degree 0. Since $\pi$ (and hence $t$) are smooth covers, by descent $L_\pi$ is concentrated in degree 0.

Now consider the composition $$ \begin{equation} X_0 \xrightarrow{\pi} \mathcal{X} \to \mathrm{Spec} \, k \end{equation} $$ and the resulting distinguished triangle $$ \pi^* L_{\mathcal{X}/k} \to L_{X_0/k} \to L_{\pi} \to \pi^* L_{\mathcal{X}/k}[1] $$ in $\mathrm{D}(X_0)$. Since $X_0$ is smooth over $k$, $L_{X_0/k}$ is concentrated in degree 0, and we just showed $L_{\pi}$ is concentrated in degree 0. Taking the long exact sequence of cohomology sheaves completes the proof.

Here's a proof assuming basic properties of the cotangent complexes of stack. At least for the proof to work I need to assume that $X_0$ is smooth over $k$ and the source/target maps $s, t: X_1 \to X_0$ of the groupoid structure are smooth (implying that $X_1$ is also smooth over $k$).

Note: when I say "concentrated in degree 0" I mean "has cohomology only in degree 0."

There's a cartesian diagram of the form $$ \begin{equation} \begin{matrix} X_1 & \xrightarrow{t} & X_0 \\ s \downarrow & & \downarrow \pi \\ X_0 & \xrightarrow{\pi} &\mathcal{X} \\ \end{matrix} \end{equation} $$

By the construction of $\mathcal{X}$, $\pi$ is a smooth cover. It follows that $$ \begin{equation} t^* L_{\pi} = L_{s} \end{equation} $$ Since $s$ is smooth by hypothesis, $L_s \simeq \Omega_{X_1/X}[0]$. So in particular $L_s$ is concentrated in degree 0. Since $\pi$ (and hence $t$) are smooth covers, by descent $L_\pi$ is concentrated in degree 0.

Now consider the composition $$ \begin{equation} X_0 \xrightarrow{\pi} \mathcal{X} \to \mathrm{Spec} \, k \end{equation} $$ and the resulting distinguished triangle $$ \pi^* L_{\mathcal{X}/k} \to L_{X_0/k} \to L_{\pi} \to \pi^* L_{\mathcal{X}/k}[1] $$ in $\mathrm{D}(X_0)$. Since $X_0$ is smooth over $k$, $L_{X_0/k}$ is concentrated in degree 0, and we just showed $L_{\pi}$ is concentrated in degree 0. Taking the long exact sequence of cohomology sheaves completes the proof.