Let $\phi=(p\vee(q\wedge r))\wedge (r\vee(p\wedge q))$ and let $\chi = (p\wedge (r\vee (p\wedge q)))\vee (r\wedge (p\vee (q\wedge r)))$. For your De Morgan lattice ${\bf M}$ we have $\phi\models_{\bf M} \chi$, but we have $\phi\not\models_{\bf L} \chi$ for the De Morgan lattice:
The complementation is defined by the self-duality that fixes $p$ and $r$.
This example is more complicated than Adam's example, but it has the property that it is purely about the underlying lattice.
The justification is this: Let ${\bf L}^*$ be ${\bf L}$ minus its top and bottom. ${\bf L}^*$ is the splitting lattice for the $p$-modular law, which is $\phi\leq \chi$. Any lattice either satisfies the $p$-modular law, or has a copy of ${\bf L}^*$ as a sublattice. Your lattice ${\bf M}$ doesn't have a sublattice isomorphic to ${\bf L}^*$, so it satisfies the $p$-modular law. On the other hand, ${\bf L}$ does not satisfy the $p$-modular law, as you can see by assigning to the variables $p, q, r$ the values indicated in the figure.
Notice that if ${\mathbf L}$ is any bounded lattice, then the ordinal sum ${\mathbf 1}+{\mathbf L}+{\mathbf L}^{\partial}+{\mathbf 1}$ has a proper De Morgan complementation satisfying all four bullet points of the problem. Any canonical example ${\mathbf M}$ would have to fail every lattice identity that failed this ordinal sum, hence every lattice identity that failed in ${\mathbf L}$. Since ${\mathbf L}$ is arbitrary, it follows that the underlying lattice of any canonical example would have to generate the variety of all lattices.