Summability for Cauchy Product Let $\sum_{m\in M}a_m$ and $\sum_{n\in N}b_n$ be summable.
Proof that the product is summable: $\sum_{(m,n)\in M\times N}a_m b_n$
Now, let $e^{sA}:=\sum_{m\in M}\frac{s^mA^m}{m!}$ and $e^{tA}:=\sum_{n\in N}\frac{t^nA^n}{n!}$ for some bounded operator $\|A\|<\infty$.
Proof that it holds: $e^{sA}e^{tA}=e^{(s+t)A}$
 A: I think so. Assume $\sum_{n = 0}^{\infty} \sum_{m = 0}^{\infty}  |A_{m,n}|$ is finite.
A) It works if $\forall m,n A_{m,n} \ge 0$ because the sum in the triangle  $\sum_{k=0}^K\sum_{l=0}^k A_{k-l,l}$ can be sandwiched between two squares $\sum_{n = 0}^{K/2} \sum_{m = 0}^{K/2}  A_{m,n}$ 
(assume $K$ is even for ease of notation)
and $\sum_{n = 0}^{K} \sum_{m = 0}^{K}  A_{m,n}$.  
B) Now, use the common trick $A_{mn} = A^+_{m,n} -  A^-_{m,n} $ and Dominated Convergence
to show the result for arbitrary $A_{mn}$
A: Denote the limit by $A:=\lim_M\sum_{m\in M}a_m$ and $B:=\lim_N\sum_{n\in N}b_n$. Then:
$$\|\sum_{m\in M}a_m\sum_{n\in N}b_n-AB\|\\
=\|(\sum_{m\in M}a_m-A+A)(\sum_{n\in N}b_n-B+B)-AB\|
\\
=\|(\sum_{m\in M}a_m-A)(\sum_{n\in N}b_n-B)+(\sum_{m\in M}a_m-A)B+A(\sum_{n\in N}b_n-B)+AB-AB\|\\
\leq\|\sum_{m\in M}a_m-A\|\cdot\|\sum_{n\in N}b_n-B\|+\|\sum_{m\in M}a_m-A\|\cdot \|B\|+\|A\|\cdot\|\sum_{n\in N}b_n-B\|\\
<\frac{\epsilon}{3}+\frac{\epsilon}{3}+\frac{\epsilon}{3}=\epsilon\quad\text{ for all } M\geq M_\epsilon\text{ and }N\geq N_\epsilon$$
Since it is summable rearrangement do not affect the sum. Thus:
$$e^{sA}e^{tA}=\sum_{m=0}^\infty\frac{s^mA^m}{m!}\sum_{n=0}^\infty\frac{t^nA^n}{n!}\\
=\sum_{l=0}^\infty\sum_{k=0}^l\frac{1}{k!}\frac{1}{(l-k)!}s^kA^kt^{(l-k)}A^{(l-k)}\\
=\sum_{l=0}^\infty\frac{1}{l!}\left(\sum_{k=0}^l\binom{l}{k}s^kt^{(l-k)}\right)A^l\\
=\sum_{l=0}^\infty\frac{1}{l!}(s+t)^lA^l=e^{(s+t)A}$$
