If $a_{n+1} \ge a_n/2$ for all $n$, then $L(A) = [-s,s]$ where $s = \sum_n a_n$.

On the other hand, if $a_{n+1} < a_n/2$ for all $n$, then $L(A)$ is a nowhere dense Cantor-type set, and you can recover $A$ from $L(A)$. 

EDIT: I don't have references, but these are not hard to prove.
As Gerald Edgar remarks, it's equivalent to consider sums $S(A) = \sum_{j \in A} a_j$ for  $A \subseteq \mathbb N$.  

In the case $a_{n+1} \ge a_n/2$, so 
$\sum_{j > n} a_j \ge a_n$, you can use a greedy approach: to make a sum $s$, 
where $0 \le S(\mathbb N)$, include $n$ in $A$ iff $S(A \cap [0\ldots n-1]) + a_n \le s$, and note that by induction $$S(A \cap [0\ldots n]) + S([n+1 \ldots \infty)) \ge s \ge S(A \cap [0\ldots n])$$

In the case $a_{n+1} < a_n/2$, consider the intervals
$J_n(A) = [S(A), S(A \cup [n+1\ldots \infty)]$ for $A \subseteq [1,\ldots, n]$,
so that $S(B) \in J_n(A)$ if $A \subseteq B \subseteq A \cup [n+1\ldots\infty)$,
and note that $J_{n+1}(A)$ and $J_{n+1}(A \cup \{n+1\})$ are disjoint closed subintervals of $J_n(A)$.