If we only consider the case that the prime $p$ in the question is odd (which amounts to considering simple groups which do not have elementary Abelian Sylow $2$-subgroups), another class of simple groups is $G = {\rm PSL}(2,q)$ when $q$ is a Mersenne prime greater than $3$.
For any odd prime divisor $p \neq q$ of $|G|$, we see that $G$ has a cyclic Sylow $p$-subgroup $P$ with $|N_{G}(P)|$ dihedral of order $2|P|$, so that $G$ has a dihedral subgroup of order $2p$ but no element of order $2p$ (note that $P$ itself need not have order $p).$ The only other odd prime divisor of $|G|$ is $q$ itself, and a Sylow $q$-subgroup $Q$ of $G$ has $|N_{G}(Q)|$ of order $\frac{q(q-1)}{2}$, which is odd. It follows that no non-identity element of $Q$ is conjugate to its inverse in $G$ ( for if it were, the conjugation could be effected within $N_{G}(Q)$, which is clearly impossible). Thus $G$ has no dihedral subgroup of order $2q$.