One thing that's interesting about classical propositional logic is the presence of a group inside it. In any Boolean ring, the XOR $\oplus$-reduct is a Boolean group.
If we take a Heyting algebra, though, as an equational theory, then I'm pretty sure we can't uniformly define any group. Uniformly meaning that we pick the same definition in terms of primitive functions in every algebra.
Boolean algebras are all Heyting algebras, so some of them contain groups, but I'm pretty sure there's no way to define a connective that's undoable in intuitionistic logic.
Is there a straightforward way to show that the theory of Heyting algebras doesn't contain a group?
The following is entirely due to Emil Jerabek at Mathoverflow; for that reason I've made this CW, and if Emil posts an answer of his own here I'll delete this one.
Your intuition is correct. Specifically, the way we prove this is by showing that the only one-variable term which the equational theory of Heyting algebras proves is an involution is the identity. This is of course nontrivial, but is a not-too-hard argument; the key phrase is "Rieger–Nishimura lattice." Urquhardt's paper Free Heyting algebras also looks like a good source, although I haven't read it yet.
If we take the above for granted, then things become quite easy. Suppose there were terms $t_*$, $t_0$, $t_-$ such that for every Heyting algebra $\mathcal{H}=(H;...)$ we have $\mathcal{H}':=(H; t_*^\mathcal{H}, t_0^\mathcal{H}, t_-^\mathcal{H})$ is a group. Then $t_-$ must be the identity (per the above) so $\mathcal{H}'$ is in fact a group of exponent $2$. Now this means that every left-multiplication function $h\mapsto g*h$ (allowing for the obvious abuse of notation) is again an involution and hence again the identity ... which means that our group $\mathcal{H}'$ is trivial.