For any $T \subseteq \mathbb N$, let
$$ A(T) = \sum_{n \in T} a_n (-1)^{B_n}$$ Now $A = A(T) + A(\mathbb N \backslash T)$ where $A(T)$ and $A(\mathbb N \backslash T)$ are independent, and so if $A$ had atoms both $A(T)$ and $A(\mathbb N \backslash T)$ would have atoms. Thus it suffices to find $T$ such that $A(T)$ has no atoms.

In particular, take an increasing sequence $T = \{t_1, t_2, \ldots\}$ such that $0 < a_{t_{j+1}} < a_{t_j}/2$. Then the function $f_T(b_1, b_2, \ldots) = \sum_{j} a_{t_j} b_j$ on $\{-1,1\}^\mathbb N$ is one-to-one. For any real $c$ there is at most one sequence of values of $B_t$, $t \in T$ that makes $A(T) = c$, and thus $P(A(T)=c) = 0$.

The answer is no.

For any $T \subseteq \mathbb N$, let
$$ A(T) = \sum_{n \in T} a_n (-1)^{B_n}$$ Now $A = A(T) + A(\mathbb N \backslash T)$ where $A(T)$ and $A(\mathbb N \backslash T)$ are independent, and so if $A$ had atoms both $A(T)$ and $A(\mathbb N \backslash T)$ would have atoms. Thus it suffices to find $T$ such that $A(T)$ has no atoms.

In particular, take an increasing sequence $T = \{t_1, t_2, \ldots\}$ such that $0 < a_{t_{j+1}} < a_{t_j}/2$. Then the function $f_T(b_1, b_2, \ldots) = \sum_{j} a_{t_j} b_j$ on $\{-1,1\}^\mathbb N$ is one-to-one. For any real $c$ there is at most one sequence of values of $B_t$, $t \in T$ that makes $A(T) = c$, and thus $P(A(T)=c) = 0$.