Is there a cyclic group $\langle G,*,e\rangle$ such that for all $ n \in \mathbb{N} - \{0,1\}$
There exists $a \in G -\{e\}$ such that $ a^n = e$ ? ,where $e$ is the neutral element.
I think this is false because cyclic groups are isomorphic to $\mathbb{Z}_n$ and there is always $ 1 < m \leq n$ such that $x^m =e$ iff $x =e$ but i am not sure.
On
If $\,|G| = k\,$ then $a^{\large k} = e.\,$ If also $\,a^{\large n} = e\,$ and $\,\gcd(n,k)=1\,$ then $a$ has order $=\color{#c00}{\bf 1}$ (since the order divides the coprimes $n,\,k),\ $ so $\ a^{\large\color{#c00}{\bf 1}}=e.\,$ Thus $\,a\neq e\,\Rightarrow\,a^n\neq 1\,$ for all $n$ coprime to $k$.
No, there is no such group $G$. If your cyclic group has order $N\in\mathbb N$, let $p$ be a prime number such that $p\nmid N$ and let $a\in G\setminus\{e\}$. Then $a^p\neq e$. That's so because, since $a\neq e$ and $p$ is prime, $a^p=e\implies p=\operatorname{ord}(a)$. But $\operatorname{ord}(a)\mid N$.