There is a pathological example that pretty much demonstrates that the existence of such an element gives no significant information about the group:

Example. Let $H$ be any finite group, and let $\pi(H)=\{p_1,\dots, p_k\}$. Now let $C$ be a cyclic group of order $p_1\cdot p_2\cdots p_k$ with generator $c$. Then $\pi(H\times C)=\pi(H)$ and $H\times C$ has an element $g=(1,c)$ of order $n=p_1\cdot p_2 \cdots p_k$, i.e $\pi(n)=\pi(H)$.

This example is also relevant to your generalized question. My hunch is that if you prescribe that $G$ is perfect, i.e. $G=G'$, then there will never exist an element of the kind you seek... But even then I'm not sure.

There is a pathological example that pretty much demonstrates that the existence of such an element gives no significant information about the group:

Example 1. Let $H$ be any finite group, and let $\pi(H)=\{p_1,\dots, p_k\}$. Now let $C$ be a cyclic group of order $p_1\cdot p_2\cdots p_k$ with generator $c$. Then $\pi(H\times C)=\pi(H)$ and $H\times C$ has an element $g=(1,c)$ of order $n=p_1\cdot p_2 \cdots p_k$, i.e $\pi(n)=\pi(H)$.

@Someone has suggested a second example which allows one to construct perfect groups with the required property (and thereby deals with my earlier remark "My hunch is that if you prescribe that $G$ is perfect, i.e. $G=G'$, then there will never exist an element of the kind you seek... But even then I'm not sure"):

Example 2. Let $S$ be any simple group and let $\pi(S)=\{p_1,\dots, p_k\}$. Now let $G=\underbrace{S\times \cdots \times S}_k$ and observe that $\pi(G)=\pi(S)$. Now let $s_i$ be an element of order $p_i$ in $S$ and observe that $g=(s_1,\dots, s_k)\in G$ has order $\pi(S)$.

Both examples are also relevant to your generalized question.