Group structure for distributive lattices
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.
This question has already been answered, but I will add a slightly different answer.
First, the question and YCor's answer both seem to assume that by a natural group structure we mean that the group is a reduct of the original structure. This means that the group operations are given as words in the original structure, as symmetric difference is given by the Boolean word $x\oplus y = (x\wedge \neg y)\vee (\neg x\vee y)$. I also make this assumption.
Any compatible relation on a structure is also a compatible relation of any reduct. (A relation $R\subseteq A^n$ is compatible on an algebraic structure $A$ if $R$ is a subalgebra of $A^n$.)
YCor's answer uses automorphisms to answer the question. The graph of an automorphism of $A$ is a compatible binary relation on $A$, so any automorphism of a distributive lattice will also be an automorphism of any reduct. Then YCor exhibits a nonidentity automorphism of a distributive lattice that fixes more than half the points, which is something that can't happen in a group.
But to me the more obvious relation to consider is the order relation $\leq$. Every nontrivial distributive lattice has a nondiscrete compatible order, while no nontrivial group has a nondiscrete compatible order. (Even nontrivial ordered groups do not have nondiscrete partial orders that are compatible with all group operations. Reason: $a\leq b$ for compatible $\leq$ implies $a^{-1}\leq b^{-1}$, which implies $a\cdot \underline{a^{-1}}\cdot b\leq a\cdot \underline{b^{-1}}\cdot b$, which implies $b\leq a$.)