Denote your grammar by $G$ and let
$$
M=\{w\in\{a,b\}^*\mid |w|_a\equiv 0\pmod 2\}
$$
where $|w|_a$ is the number of $a$s in $w$. In other words, $M$ is the set of all strings of $a$s and $b$s with an even number of $a$s. We claim $L(G)=M$.
First, it is clear that any word in $L(G)$ must have an even number of $a$s, so $L(G)\subseteq M$. All we now have to do is show $M\subseteq L(G)$, in other words, that every string in $M$ can be generated by $G$. We'll show this by induction on the number, $n$, of $a$s in a word.
If $n=0$, we can generate strings consisting only of $i\ge 0\ b$s by one of two derivations:
$$\begin{align}
S &\Rightarrow A\Rightarrow \epsilon \\
S &\Rightarrow A\Rightarrow bA\stackrel{*}{\Rightarrow} b^i
\end{align}$$
So we can generate all strings with no $a$s.
If $n=2$, any string in $M$ with exactly two $a$s must have the form $b^iab^jab^k$ with $i, j, k\ge 0$. Write such a string as $(b^i)(ab^ja)(b^k)$. We've seen that we can generate the $b^i$ and $b^k$ portions from $A$. We can also generate the $ab^ja$ part from $A$:
$$
A\Rightarrow aAa\stackrel{*}{\Rightarrow} ab^ja
$$
so we can generate $b^iab^jab^k$ by
$$
S\Rightarrow ASA\Rightarrow AAA\stackrel{*}{\Rightarrow} b^iAA\stackrel{*}{\Rightarrow}b^iab^jaA\stackrel{*}{\Rightarrow}b^iab^jab^k
$$
and it's clear that we can continue this process for even $n>2$, simply by starting with the requisite number of $A$s. In short, we've shown that any word in $M$ can be generated by the grammar $G$, and this, combined with the first part, shows that $L(G)=M$.
Of course, $M$ is regular, if for no other reason than it is denoted by the regular expression $(ab^*a+b)^*$. Also, it would also be easy to construct a finite automaton for $L$ as well.
