No:

If it's natural, it should be invariant under the automorphism group of the original lattice.

let $X$ be the free distributive lattice on 2 generators $x,y$: it has 6 elements, $$0\quad<\quad x\wedge y \quad<\quad \stackrel{x}{_y}\quad<\quad x\vee y\quad<\quad 1 $$
with $x,y$ not comparable. It has an automorphism exchanging $x$ and $y$, and fixing the other elements.

But no group of order 6 has no automorphism with this property (if $G$ is a finite group and an automorphism fixes $>|G|/2$ elements, it's identity, just because the set of fixed points is a subgroup).

Initial answer: no for arbitrary finite lattices (the following example is not distributive).

Consider the lattice of subgroups of the Klein group $C_2^2$. It has cardinal 5 (0, the whole plane and 3 lines), and the automorphism group (of order 6) has two fixed points and a 3-element orbit.

Now a group structure on 5 elements is cyclic of order 5 and its automorphism group is cyclic of order 4, so a group structure on 5 elements cannot be preserved by the original automorphism group of order 6.

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