I think the nice added conjecture goes to the core of the problem, nevertheless it has to be modified because it doesn't hold as it is.
Consider for instance an entire function
$f(x):=\sum_{n=0}^\infty a_n x^n$ with
$$|a_n|=3^{-n^2}.$$
then $f$ is unbounded on $\mathbb{R}$ since we have, for all $a\in\mathbb{N}:$
$$\left|\ f(3^{2a})\right| \geq 3^{a^2} -\sum_{n\neq a} 3^{n(2a-n)}\geq 3^{a^2}\left( 1-2\sum_{k>0 }3^{-k^2}\right)\ge \frac{3^{a^2}}{4}.$$
I may as well write down how I've been attacking this, although I don't have a solution yet. Sequences $(p_i)$ of nonnegative integers with sum $r$ are in bijection with weakly increasing $r$-tuples $(n_1, n_2, \ldots, n_r)$ of positive integers. Specifically, given the sequence $(p_0, p_1, p_2, \ldots)$, form the sequence of partial sums $(q_1, q_2, \ldots)$ given by $q_i:=p_0+p_1+\ldots+p_{i-1}$. Let $n_j$ be the minimal $i$ for which $q_i \geq j$. For example, $(1,0,0,1,0,0,2,0,0,\ldots)$ corresponds to $(1, 4, 7,7)$. So, we want to compute the sum over all weakly increasing $r$-tuples and prove it is equal to $1/r!$.
For each weakly increasing $r$-tuple, let us sum instead over all $r!$ permutations of the $r$-tuple. So we can view our sum as being over all $r$-tuples of nonnegative integers, and we want to prove now that the sum is $1$. One difficulty is that some $r$-tuples will appear more than once. For example, $(1,4,7,7)$ will appear twice, because the permutation which switches the $7$'s stabilizes this quadruple. It turns out that the multiplicity of an $r$-tuples is precisely $\prod (p_i)!$. So, what we want to show is that
$$\sum_{n_1, n_2, \ldots, n_r \geq 0} \frac{1}{\prod (p_i)! \prod \binom{p_{k-1}+p_k+k}{k}} =1$$
Now, when none of the $n_i$ are equal to each other, and when none of them differ by $1$, the summand is
$$\frac{1}{n_1(n_1+1)n_2(n_2+1) \cdots n_r(n_r+1)} = \left( \frac{1}{n_1} - \frac{1}{n_1+1} \right) \cdot \left( \frac{1}{n_2} - \frac{1}{n_2+1} \right) \cdots \left( \frac{1}{n_r} - \frac{1}{n_r+1} \right).$$
This is set up beautifully to telescope. If I could just find a similar nice description for when the $n_i$ collide or are adjacent ...
Best Answer
It seems to me that your series is related to Graf's addition theorem. There, instead of the power of $(B/A)^\lambda$, there is a trigonometric function. However, writing a complex exponential as $e^{i\lambda\theta} = (B/A)^\lambda$ could give you the answer on the complex unit circle $|B/A|=1$. Extending away from the unit circle by analyticity may give you the full answer. The main thing that's odd about your formula is that it has $(B/A)^{|\lambda|}$ instead of $(B/A)^\lambda$ in Graf's addition theorem. But perhaps that can be fixed using some reflections of the order of the Bessel functions.