Define the function $$\exp(z) = \sum_{k=0}^\infty \frac{z^k}{k!},$$ and consider the product $$\exp(z) \exp(-z) = \sum_{k=0}^\infty \sum_{m=0}^\infty \frac{(-1)^m z^{k+m}}{k! m!}.$$ If we collect like powers of $z$; i.e., letting $n = k+m$, this double sum is equivalent to $$\exp(z) \exp(-z) = \sum_{n=0}^\infty \sum_{m=0}^n \frac{(-1)^m}{m! (n-m)!} z^n = \sum_{n = 0}^\infty c_n \frac{z^n}{n!},$$ where $$c_n = \sum_{m=0}^n (-1)^m \binom{n}{m}.$$ Therefore, it suffices to show that $c_0 = 1$ and $c_n = 0$ for all $n \in \mathbb Z^+$. The zero case is trivial. The positive case is a direct corollary of the binomial theorem: $$(x+y)^n = \sum_{m=0}^n \binom{n}{m}x^{n-m} y^m,$$ with the choice $x = 1, y = -1$: $$0^n = \sum_{m=0}^n \binom{n}{m} (-1)^m = c_n.$$ Therefore, $$\exp(z)\exp(-z) = 1.$$
It is worth noting that nowhere in the above proof do we make any claims that $\exp(z)$ is a function whose value is some constant raised to the power of $z$; i.e., we don't assume $\exp(z) = e^z$, and thus we do not use the multiplicativity of $\exp$. All that is being done here is to show that this particular power series satisfies the desired property.
Maybe a different notational framework presenting the same ideas will help you understand the proof.
We begin by introducing introducing $\varepsilon_1,\varepsilon_2,\varepsilon_3$ and associated $N_1,N_2,N_3$ based on the facts that $B_n \to B$ and $A_n \to A$ absolutely. We will specify how these relate to the global $\varepsilon$ in the statement we want to prove later. For now, we do the following:
Find $N_1$ with $|B_n-B|<\varepsilon_1$ for $n \geq N_1$.
Find $N_2$ with $|a_n|<\varepsilon_2$ for $n \geq N_2$.
Find $N_3$ with $|A_n-A|<\varepsilon_3$ for $n \geq N_3$.
Algebraically write $C_n=A_n B + \sum_{i=0}^n a_{n-i}(B_i-B)$. Note that this still amounts to working with the same sequence of partial sums as you were in the first place. In other words we are only rearranging the terms in the representation of $C_n$ as a sum, we are not changing what the values of $C_n$ actually are.
Then
$$|C_n-AB| \leq |A_n-A||B| + \left | \sum_{i=0}^n a_{n-i}(B_i-B) \right |.$$
The first term is less than $\varepsilon_3 |B|$ for $n \geq N_3$.
In the second term, we demand $n$ be large enough that $n-N_1 \geq N_2$, and then split the sum at $i=N_1$:
$$|C_n-AB| \leq \varepsilon_3 |B| + \sum_{i=0}^{N_1-1} |a_{n-i}||B_i-B| + \sum_{i=N_1}^n |a_{n-i}||B_i-B|.$$
The terms of the first sum are less than $\varepsilon_2 |B_i-B|$. The terms of the second sum are less than $\varepsilon_1 |a_{n-i}|$. So we have an estimate of
$$|C_n-AB| \leq \varepsilon_3 |B| + \varepsilon_2 \sum_{i=0}^{N_1-1} |B_i-B| + \varepsilon_1 \sum_{i=N_1}^n |a_{n-i}|.$$
So now the question is how do we select these three epsilons to get an overall estimate less than $\varepsilon$? The usual idea is to just make the bounds on the pieces be the same. So you want $\varepsilon_3<\frac{\varepsilon}{3|B|}$, $\varepsilon_2<\frac{\varepsilon}{3 \sum_{i=0}^{N_1-1} |B_i-B|}$ and $\varepsilon_1<\frac{\varepsilon}{3 \sum_{i=N_1}^n |a_{n-i}|}$.
As is this is not OK because there is a residual dependence on $N_1$ in the definition of $\varepsilon_1$, which means it is not clear that such a $N_1$ necessarily exists. However, this dependence can be removed: just impose $\varepsilon_1<\frac{\varepsilon}{3 \sum_{i=0}^\infty |a_i|}$ instead. This only makes the denominator bigger than it would be for any particular values of $n$ and $N_1$. This step is crucial, because until we specify what $\varepsilon_1$ is, we don't know what $N_1$ is, while $N_1$ is also used in the second term. It is also the only place where the absolute convergence of $A_n$ to $A$ is actually used.
Having done that, $N_1$ is now fixed (since $N_1$ was defined in terms of $\varepsilon_1$ through the definition of $B_n \to B$). So you can just leave $\varepsilon_2<\frac{\varepsilon}{3\sum_{i=0}^{N_1-1} |B_i-B|}$ as it is and get the desired inequality, which is
$$|C_n-AB| \leq \varepsilon$$
if $n \geq \max \{ N_3,N_1+N_2 \}$.
Best Answer
this kind of product is called the Cauchy product of two series :) You got the hang of it for the first few terms, now, by induction, it generalizes to:
$\left(\displaystyle\sum_{n=0}^{+\infty} a_n x^n\right) \cdot \left(\displaystyle\sum_{n=0}^{+\infty} b_n x^n\right) = \displaystyle\sum_{n=0}^{+\infty} c_n x_n$, where each of the $c_n$ is defined as $c_n = \displaystyle\sum_{j = 0}^{n} a_j b_{n-j}$.
This coincides with the first few terms you computed :). You can see more on the wikipedia page on Cauchy product.
PS: try to LaTeX your question next time, it makes it clearer to read :)