Just for fun, here's a purely abstract way of seeing neither of these can happen:

The functors you wrote down both preserve weak equivalences, so are morphisms of relative categories, and induce adjunctions of the associated $\left(\infty,1\right)$-categories, which are $\infty$-groupoids and $1$-groupoids (which is actually a $(2,1)$-category) respectively. The adjunction is precisely the one exhibiting $1$-groupoids as a reflective subcategory of $\infty$-groupoids. The full and faithful inclusion is modeled by $N$. The $\left(\infty,1\right)$-category of $\infty$-groupoids is freely generated by the (any) contractible $\infty$-groupoid (i.e. the one point space), (i.e. it's the free colimit cocompletion of the terminal category). Since the $\left(2,1\right)$-category of groupoids is cocomplete, and contains the terminal groupoid, if the inclusion preserved colimits, it would have to be essentially surjective:

If $X$ is any $\infty$-groupoid, you can write $X$ as the colimit of the terminal functor $X \to \infty\mbox{-}\mathbf{Gpd}$ (the constant functor with value the point) (this is basically just the Yoneda Lemma for $\infty$-categories), and this functor (obviously) factors through the inclusion $\mathbf{Gpd} \hookrightarrow \infty\mbox{-}\mathbf{Gpd}$.

If the left adjoint $\tau_1$ of this inclusion (which is modeled by $\Pi_1$) preserved even **finite** limits, then this would exhibit the $(2,1)$-category of groupoids as a left-exact localization of $\infty\mbox{-}\mathbf{Gpd}=\mathbf{Psh}_\infty\left(*\right),$ i.e. it would exhibit $\mathbf{Gpd}$ as an $\infty$-topos. This means there would exist a unique Grothendieck topology $J$ on the terminal category such that $\mathbf{Gpd}$ sat somewhere between $\mathbf{Sh}_\infty\left(*,J\right)$ and its hypercompletion $\mathbf{HSh}_\infty\left(*,J\right).$ But, there is a unique Grothendieck topology on the terminal category (the trivial one!), so we'd have $$\mathbf{HSh}_\infty\left(*,J\right)=\mathbf{Sh}_\infty\left(*,J\right)=\mathbf{Psh}_\infty\left(*\right).$$ So we'd have to have $\mathbf{Gpd}\simeq \mathbf{Psh}_\infty\left(*\right)=\infty\mbox{-}\mathbf{Gpd}$ which is clearly nonsense.