Generalization of a theorem of Øystein Ore in group theory

This is not a proper answer, but is a bit too long for a comment. You should look at the following paper:

Groups and Lattices by P.P. Palfy

First, he states a stronger version of Ore's result:

Theorem: Let $G$ be any group. The subgroup lattice $\mathcal{L}(G)$ is distributive if and only if the group $G$ is locally cyclic. (i.e. every finitely generated subgroup of $G$ is cyclic.)

Second he states a theorem (proved by various people at various times, sometimes erroneously):

Theorem If $D$ is a finite distributive lattice, then $D$ is isomorphic (as a lattice) to the lattice of normal subgroups of some finite group $G$.

Finally the result that may be of most interest to you:

Corollary 5.3 For every finite distributive lattice $D$, there exists a finite group $G$ such that, for $H=\{(g,g) \mid g\in G\}$, then the lattice of subgroup inclusions $\mathcal{L}(H\subset (G\times G))$ is isomorphic to $D$.

Perhaps you know all this already...

The answer of the main question is yes. Let $G$ be a finite group and $H$ a subgroup.

Definition: The group $G$ is called $H$-cyclic if $\exists g \in G$ such that $\langle H,g \rangle = G$.
Note that: $\langle H,g \rangle = G \Leftrightarrow \langle Hg \rangle = G$.
Ore's theorem for intervals (1938): If the interval $[H,G]$ is distributive, then $G$ is $H$-cyclic.
proof: see here, Theorem 7 p269.

The following is a new proof.

Lemma 0: The top interval of a distributive lattice, is boolean.
proof: see here.
Lemma 1: If $H$ is a maximal subgroup then $G$ is $H$-cyclic.
proof: immediate.
Lemma 2: Let $[K,G]$ be the top interval of $[H,G]$. If $G$ is $K$-cyclic then it is $H$-cyclic.
proof: straightforward.

So the proof of Ore's theorem for intervals reduces to the following Theorem.

Theorem: If the interval $[H,G]$ is boolean, then $G$ is $H$-cyclic.
proof: Let $M$ be a coatom of $[H,G]$, and $M^{\complement}$ its complement. By induction on the rank of the lattice (and Lemma 1), we can assume $M$ and $M^{\complement}$ both $H$-cyclic, i.e. there are $a, b \in G$ such that $\langle H,a \rangle = M$ and $\langle H,b \rangle = M^{\complement}$. Let $g=a b$ then $a=g b^{-1}$ and $b=a^{-1}g$, so $\langle H,a,g \rangle = \langle H,g,b \rangle = \langle H,a,b \rangle = M \vee M^{\complement} = G$.
Now, $\langle H,g \rangle = \langle H,g \rangle \vee H = \langle H,g \rangle \vee (M \wedge M^{\complement})$, but by distributivity $\langle H,g \rangle \vee (M \wedge M^{\complement}) = (\langle H,g \rangle \vee M \rangle) \wedge (\langle H,g \rangle \vee M^{\complement} \rangle)$.
So $ \langle H,g \rangle = \langle H,a,g \rangle \wedge \langle H,g,b \rangle = G$. The result follows. $\square$

The converse of Ore's theorem is false because $\langle S_2,(1234) \rangle = S_4$ but $[S_2, S_4]$ is not distributive.