Here is an answer to question (2), strongly inspired by Dave Benson's comment:
Theorem Let $A$ be any ring and let $M$ be a finite length $A$-module. Then the lattice of $A$-submodules is distributive iff $M$ does not have a subquotient of the form $S^2$, for $S$ simple.
Proof First, suppose that $M$ has an $S^2$ subquotient, say $X \subset Y \subseteq M$ with $Y/X \cong S^2$. Then the interval $[X,Y]$ in the submodule lattice of $M$ is isomorphic to the submodule lattice of $S^2$. In particular, the submodule lattice of $S^2$ contains $S \oplus \boldsymbol{0}$, $\boldsymbol{0} \oplus S$ and $\Delta:= \{ (s,s) : s \in S \}$. Then $\{ \boldsymbol{0}, S \oplus \boldsymbol{0}, \boldsymbol{0} \oplus S, \Delta, S^2 \}$ form a copy of the $M_3$ diamond lattice, which is not distributive.
Conversely, suppose that the submodule lattice of $M$ is not distributive. Then it must contain either a copy of $M_3$ or $N_5$. But the submodule lattice is modular, so it cannot contain $N_5$, so it must contain $M_3$. In other words, there are submodules $X \subset W_1, W_2, W_3 \subset Y$ with $W_i \cap W_j = X$ and $W_i + W_j = Y$ for all $i \neq j$. Put $Z = Y/X$ and $V_i = W_i/X$. So $Z$ has submodules $V_i$ such that $W_i \cap W_j = \{ 0 \}$ and $W_i + W_j = Y/X$ for $i \neq j$. Then we have
$$V_2 \cong Z/V_1 \cong V_3 \cong Z/V_2 \cong V_1 \cong Z/V_3$$
so $V_1 \cong V_2 \cong V_3$ and $Z \cong V_1^2$. Let $S$ be a simple quotient of $V$, then $S^2$ is a quotient of $Z$ and is a subquotient of $Y/X$. $\square$
Corollary: The submodule lattice of $M_1 \oplus M_2 \oplus \cdots \oplus M_r$ is distributive iff (1) the submodule lattice of each $M_i$ is individually distributive and (2) for $i \neq j$, the modules $M_i$ and $M_j$ have no common simple subquotient.
Now, in the preprojective Dynkin case, every simple module is a subquotient of every indecomposable projective module. So this simplifies to
Corollary: In the original Dynkin case that the OP asked about, if a projective module has distributive submodule lattice then it is indecomposable.
Here is the answer to question (1). The first thing to know is that the premise of the question is wrong: Indecomposable projectives do not always have a distributive submodule lattice. Rather, $P_i$ has a distributive submodule lattice if and only if the weight $\omega_i$ is minuscule. (Because we are in simply-laced type, "minuscule" and "co-minscule" are synonyms.) In this case, the submodule lattice of $P_i$ is isomorphic to the lattice of order ideals in the heap $H(w^i)$.
The minuscule weights have been classified: See Table 1 in Thomas-Yong and ignore the rows for types $B$ and $C$. The corresponding heaps are also well known; see the Figures in Section 2.2 of the Thomas-Yong paper.
I will now explain how to piece these results together, using facts from Baumann and Kamnitzer and Dranowski, Elek, Kamnitzer and Morton-Ferguson.
We use the usual Coxeter notations: $\alpha_i$ are the simple roots, $s_i$ are the simple reflections and $\omega_i$ are the fundamental weights.
To every weight $\gamma$, Baumann and Kamnitzer define a preprojective module $N(\gamma)$, such that $N(\omega_i) = P_i$ (see Prop 10). Here are the key facts about $N(\gamma)$:
- If $\gamma$ is antidominant, then $N(\gamma) = 0$.
- If $s_j \gamma = \gamma + c \alpha_j$ for some $c \geq 0$, then we have a short exact sequence $0 \to N(\gamma) \to N(s_j \gamma) \to S_j^c \to 0$.
- As a corollary of the proceeding two statements, let $\gamma'$ be the unique antidominant weight in the same orbit as $\gamma$ and let $\gamma - \gamma' = \sum d_i \alpha_i$. Then $N(\gamma)$ has dimension $(d_1, d_2, \ldots, d_n)$.
- Let $\gamma$ and $\gamma'$ be as in the previous bullet point, and let $\gamma' = - \sum c_i \omega_i$. Then $N(\gamma)$ is uniquely characterized as the only preprojective module with dimension $(d_1, d_2, \ldots, d_n)$ and socle of dimension $(c_1, c_2, \ldots, c_n)$.
Now, let $\omega'_i$ be the unique fundamental weight so that $-\omega'_i$ and $\omega_i$ are in the same orbit. Let $w^i$ be the Coxeter group element of minimal length such that $w^i \omega_i = - \omega'_i$ and let $s_{j_1} s_{j_2} \cdots s_{j_L}$ be a reduced word for $w^i$. Put $\gamma(k) = s_{j_k} \cdots s_{j_2} s_{j_1} (- \omega'_i)$, so $\gamma(0) = - \omega'_i$, $\gamma(L) = \omega_i$ and $\gamma(k) = \gamma(k-1) + c_k \alpha_{i_k}$ for some integer $c_k$; it turns out (proof omitted) that $c_k>0$. So the modules $N(\gamma(k))$ form a filtration of $N(\omega_i) = P_i$, with subquotients $S_{j_k}^{c_k}$.
In particular, if any $c_k$ is $\geq 2$, then the submodule lattice of $P_i$ is not distributive.
If all the $c_k$ are $1$, then $w^i$ is what is called $\omega_i$-minuscule (see Definition 2.4 in Dranowski, Elek, Kamnitzer and Morton-Ferguson), and is therefore minuscule, and the results of that paper apply. I'll summarize those results now.
We define a poset $H(w^i)$, called the heap of $w^i$, as follows: The ground set of $H(w^i)$ is $[L]$. The order relation is the transitive closure of the following: $a \prec b$ if $a<b$ and $s_{j_a} s_{j_b} \neq s_{j_b} s_{j_a}$. We have the following:
$H(w^i)$ indexes a basis for $\mathbb{C} H(w^i)$. To understand the maps in $\mathbb{C} H(w^i)$, take the Hasse diagram of $H(w^i)$ (see the Thomas-Yong paper) and place $1$'s and $-1$'s on the edges in order to make the preprojective relations hold.
Order ideals of $H(w^i)$ correspond to submodules of $\mathbb{C} H(w^i)$. These also correspond to the weak order interval $[e, w^i]$, and to roots in the $W$ orbit of $\omega_i$.
The submodule lattice of $\mathbb{C}H(w^i)$ is the lattice of order ideals in $H(w^i)$.
Total orders of $H(w^i)$ correspond to reduced words for $w^i$, which correspond to composition series of $P_i$.
In type $A_n$, $H(w^i)$ is the product of an $i$-vertex chain and an $(n+1-i)$ vertex chain. Order ideals correspond to partitions fitting in an $i \times (n+1-i)$ box, and the submodule lattice is this part of Young's lattice. Total orders on $H(w^i)$ are standard Young tableaux of shape $i \times (n+1-i)$.
