Denote $A_{\infty}=\underset{n\in\mathbb{N}}{\bigcup}A_{n}$. For any finite group $H$, the group $A_{\infty}\times H$ have a composition series.
I've shown that $A_{\infty}$ is simple, therefore it's composition series is $A_{\infty}\vartriangleright\left\{ \text{Id}\right\} $. I know that $H$ have some composition series because it's finite. I think the composition series for $A_{\infty}\times H$ would be something like: $\left(A_{\infty}\times H_{0}\right)\vartriangleright\left(\left\{ Id\right\} \times H_{0}\right)\vartriangleright\left(\left\{ Id\right\} \times H_{1}\right)\vartriangleright\left(\left\{ Id\right\} \times H_{2}\right)\vartriangleright\dots\vartriangleright\left(\left\{ Id\right\} \times\left\{ e\right\} \right)$ but I'm not sure how to prove that $\left(A_{\infty}\times H_{0}\right)/\left(\left\{ Id\right\} \times H_{0}\right)$ is indeed simple.
$\newcommand{\norm}{\trianglelefteq}$In general, if $G, H$ are groups, and $N \norm G$, $M \norm H$, then $N \times M \norm G \times H$, and $$ \frac{G \times H}{N \times M} \cong \frac{G}{N} \times \frac{H}{M}. $$ This you see by checking that the map \begin{align} f :\ &G \times H \to \frac{G}{N} \times \frac{H}{M}\\ &(g, h) \mapsto (g N, h M) \end{align} is a surjective homomorphism whose kernel is $N \times M$, and then applying the first isomorphism theorem.
In your case, then, $$ \frac{A_{\infty}\times H_{0}}{ \left\{Id\right\} \times H_{0}} \cong A_{\infty} $$ is simple.