What property of modules ensures that module lattice is uniquely complemented? Background:

If a module $M$ is semisimple then every submodule $N \subseteq M$ is
  a direct summand.

In other terms, there exists a submodule $H \subseteq M$ such that $N \cap H = \{0\}$ and $N+H = M$. 
Given the lattice of submodules $L(M)$ (where the infimum is $\cap$ and the supremum is $+$), we may consider what property can we impose on the module $M$ so that $L(M)$ is complemented (for each $N$ there exists $H$ such that $N \cap H = \{0\}$ and $N+H = M$). In particular, the lattice of submodules of a semisimple module is complemented iff the module is semisimple. 
One can further impose the complement to be unique. 
Question:
What property do I need on $M$ so that $L(M)$ is uniquely complemented?
Examples:
For instance, $\mathbb{Z}_2 \oplus \mathbb{Z}_2$ has a lattice of submodules given by:
\begin{matrix}
 && \mathbb{Z}_2 \oplus \mathbb{Z}_2 & \\
&\huge\diagup & \huge| & \huge\diagdown \\
(0,1) && (1,0) && (1,1) \\
&\huge\diagdown & \huge| & \huge\diagup \\
&& 0
\end{matrix}
so we don't have the unique complement property. On the other hand, $\mathbb{Z}_2 \oplus \mathbb{Z}_3$ has a lattice of submodules given by:
\begin{matrix}
 && \mathbb{Z}_2 \oplus \mathbb{Z}_3 & \\
&\huge\diagup && \huge\diagdown  \\
(1,0) &&& (0,1) \\
&\huge\diagdown && \huge\diagup  \\
&& 0
\end{matrix}
which is uniquely complemented.
 A: You need the module to be distributive which, for a semisimple module, amounts to being square-free in terms of simple submodules. (I mean that no two distinct simple submodules are isomorphic.)
If a semisimple module isn't square-free, then you can construct the diamond lattice $M_3$ as you did in your first example, and that makes it nondistributive.
If the semisimple module is square-free, then the simple submodules that make up that square-free factorization actually have to be unique within the module. Since complements are obviously isomorphic, then they must be made up of the same simple submodules, but since those are unique, the complements are actually equal.
You will probably find this paper of Dilworth's very informative.
A: Reusing the ideas of @rschwieb (checking details for my own reference)
Lemma
Let $M$ be a semisimple module. The following statements are equivalent:
$L(M)$ is a distributive lattice.
$M$ is square-free 
$L(M)$ is uniquely complemented. 
Proof
We proof circularly the implications:
Distributive $\implies$ square free
If a semi-simple module isn't square-free, then we can construct a diamond lattice  as in the example above and that makes it non-distributive. 
Indeed, if it is not square free it means that there are two simple submodules $S_1\cong S_2$ where $f$ is an isomorphism between the two. In this case, we may build the submodule $S_1 \times f(S_2) = \{(s,f(s)):s \in S_1\}$. It is not difficult to prove from here that in fact $S_1,S_2$ have as complement $S_1 \times f(S_2)$ (one can do it by elements). And therefore we would have completed a diamond lattice. Therefore, the lattice cannot be distributive. 
Square free $\implies$ uniquely complemented
Given submodule $N$. The possible complements are isomorphic. If $N_1,N_2$ are two distinct complements, then $M = N \oplus N_1 = N \oplus N_2$. Then, taking quotient we would have $M/N \cong N_1 \cong N_2$ (to see this apply first isomorphism theorem to the projection going out of the sum).
Since module homomorphisms direct and inverse image preserve submodules, simple submodules are invariant under these homomorphisms which means that there is a bijection between simple submodules of $N_1$ and $N_2$, taking one simple submodule to an isomorphic pair. But since $N$ is square free, bijection is actually and identity, so that $N_1,N_2$ have the same simple submodules. 
Since, $N_1,N_2$ are semi-simple, they are the sum of their simple submodules and therefore $N_1 = N_2$. So complements of $N$ are equal. 
Uniquely complemented $\implies$ Distributive
This appears to be a general proposition whose proof can be found in "Lattices and Ordered Algebraic Structures" by T.S. Blyth and is attributed to Von Neumann.
