I have a few observations about this question,
but only time today to write down one of them.
For this I will write $H\cap K$ for the intersection
of two subgroups (just as everyone else does), and write
$H+K$ for the join of the subgroups.

**Theorem.**
Let $G$ be a group, and let $(P_1,Q_1)$ and
$(P_2,Q_2)$ be two pairs of complementary normal
subgroups (a.k.a. pairs of complementary
direct factor subgroups of $G$).
If $P = P_1\cap P_2$ and $Q = Q_1+Q_2$, then

 1. $[G,G]\subseteq P+Q$.
 2. $[P\cap Q,P+Q] = \{1\}$.

Therefore, if $G$ is any (finite) centerless, perfect group, then
$G$ is a $\mathcal D$-group.

**Proof:**
For the first item,
$$
\begin{array}{rl}
[G,G]&=[P_1+Q_1,P_2+Q_2]\\
&=[P_1,P_2]+[P_1,Q_2]+[Q_1,P_2]+[Q_1,Q_2]\\
&\leq [P_1,P_2]+Q\\
&\leq (P_1\cap P_2) + Q = P+Q.
\end{array}
$$
Here I am using the additivity of the commutator, the fact that
$[H,K]\leq H\cap K$, and the fact that $Q_1, Q_2\leq Q$.

For the second item,
$[P,Q_1] \leq P\cap Q_1 \leq P_1\cap Q_1 = \{1\}$.
Similarly
$[P,Q_2] = \{1\}$. By the additivity of the commutator,
$[P,Q]=[P,Q_1+Q_2]=[P,Q_1]+[P,Q_2]=\{1\}$. Now let $Z=P\cap Q$, which is $\leq P$ or $Q$. From the last two sentences and the monotonicity
of the commutator in each variable we deduce
$[Z,Q]\leq [P,Q] = \{1\}$ and $[Z,P]\leq [Q,P]=[P,Q]=\{1\}$,
so by additivity we get 
$$[P\cap Q,P+Q]=[Z,P+Q]=[Z,P]+[Z,Q]=\{1\}.$$ 
This
is the assertion to be proved.

For the final sentence of the proof, let $G$
be a perfect group ($[G,G]=G$) that is also a
centerless group ($[G,N]=\{1\}$ implies $N=\{1\}$).
Using the perfectness of $G$, 
the first item of the theorem can be written $G\subseteq P+Q$.
Using this (i.e. $G=P+Q$), the second item can be written
$[P\cap Q,G]=\{1\}$, or $P\cap Q\leq Z(G)$. Using the
centerlessness of $G$ we get $P\cap Q=\{1\}$.
Altogether we obtain that $P=P_1\cap P_2$ and $Q=Q_1+Q_2$ 
are complementary
normal subgroups of $G$.
This shows that the collection of
factor congruences is closed under $\cap$ and $+$,
so $G$ is a $\mathcal D$-group \\\\\\

[One can go a bit further and show that the lattice of factor
subgroups of a perfect, centerless group is a 
complemented **distributive** sublattice
of ${\mathcal N}(G)$.]