Let $\mathcal{G} = G_1 \rightrightarrows G_0$ be a Lie Groupoid (although I am also interested in groupoids internal to other sites), the stack associated to $\mathcal{G}$, which is sometimes denoted $B \mathcal{G}$ is defined as the projection from the category of principal $\mathcal{G}$-bundles to $\mathsf{Man}$. However it is also well known that we can use the functor $r_1: \mathsf{LieGpd} \hookrightarrow \mathsf{Pre}(\mathsf{Man}, \mathsf{Gpd})$, defined as $$M \mapsto \left( C^\infty(M, G_1) \rightrightarrows C^\infty(M, G_0) \right)$$ and that the stackification of this presheaf of groupoids is equivalent to $B \mathcal{G}$.
However as Chris Schommer-Pries pointed out, $r_1(G \rightrightarrows *)$ is not a stack on $\mathsf{Man}$. It is however a stack on $\mathsf{Cart}$ the site of cartesian spaces. This can be seen by showing that if we let $\widehat{r}G = r_1(G \rightrightarrows *)$, then for $U \in \mathsf{Cart}$ equipped with a differentiably good open cover $\{ U_i \to U \}$, the canonical map
$$\widehat{r}G(U) \to \text{holim} \left( \prod_i \widehat{r}G(U_i) \rightrightarrows \prod_{i,j} \widehat{r}G(U_{ij}) \mathrel{\substack{\textstyle\rightarrow\\[-0.6ex] \textstyle\rightarrow \\[-0.6ex] \textstyle\rightarrow}} \prod_{i,j,k} \widehat{r}G(U_{ijk}) \right)$$ is an equivalence of groupoids. One can see this by using a specific model of $\text{holim}$ from Hollander's Homotopy Theory of Stacks Corollary 2.11, and then recognizing the right hand side as the groupoid of Principal $G$-bundles on $U$. Since $U$ is a Cartesian space, this groupoid will be connected and this will prove that the map is an equivalence.
My question then is the following:
Is there a necessary and sufficient condition that one could put on a Lie groupoid $\mathcal{G}$ such that $r_1(\mathcal{G})$ is already a stack on $\mathsf{Cart}$? I suspect that it should be any Lie groupoid, we would need to follow the same logic as above using principal groupoid bundles, i.e. building principal groupoid bundles from local data. Perhaps this has been done somewhere?