Yes

More generally, $\mu(1,G)=\pm 1$ iff $G$ is cyclic of square-free order (iff $\mathcal{L}(G)$ is boolean).

Théorème 3.1. of the paper "Fonction de Möbius d'un groupe fini et anneau de Burnside" (1984) by Kratzer and Thévenaz (available here) states the following (with $n_0$ the square-free part of $n$): $$\mu(1,G) \in \frac{|G|}{|G:G'|_0} \mathbb{Z}$$

But if $\mu(1,G)=\pm 1$ then $|G|= |G:G'|_0$, and so $G'=1$. It follows that $G$ is abelian with $|G|$ square-free, so $G$ is cyclic of square-free order and $\mathcal{L}(G)$ is boolean. The converse is immediate.

Yes, and a much more general statement is true.

First, note that if $\mathcal{L}(G)$ is an Eulerian lattice then $\mu(1,G)=\pm 1$.

Theorem: $\mu(1,G)=\pm 1$ iff $G$ is cyclic of square-free order iff $\mathcal{L}(G)$ is boolean.

Proof: Théorème 3.1. of the paper "Fonction de Möbius d'un groupe fini et anneau de Burnside" (1984) by Kratzer and Thévenaz (available here) states the following (with $n_0$ the square-free part of $n$): $$\mu(1,G) \in \frac{|G|}{|G:G'|_0} \mathbb{Z}$$

But if $\mu(1,G)=\pm 1$ then $|G|= |G:G'|_0$, and so $G'=1$. It follows that $G$ is abelian with $|G|$ square-free, so $G$ is cyclic of square-free order and $\mathcal{L}(G)$ is boolean. The converse comes from a theorem of Ore stating that $G$ is cyclic iff $\mathcal{L}(G)$ is distributive. $\square$