Context free grammar where number of a's is larger than number of b's in any prefix of the string
In short, take your grammar for  the language having a greater or equal number of $a$'s in every prefix, but rename the startsymbol to $P$, i.e.
$$
 P \to aP \mid aPbP \mid \varepsilon.
$$
Then simply note that your condition requires a word to start with an $a$, hence the following will do the job:
$$
 S \to aP.
$$
For if $S \to w$ then $w = auv$ with $|u|_a \le |u|_b$, where $|u|_a$ denotes the number of $a$'s in $u$ (and analog for $|u|_b$), hence for the prefixes of $w$ the assertion follows. And conversely it works the same.

For your given grammar, this is correct (but I leave my answer as it seems to be simpler). For a proof note that if the produced word contains at least one $b$, then it must have at least two $a$'s before it, and this is what your production $aSAbA$ takes care of, where from $A$ we can just derive words having at most as many $a$'s as $b$'s in every prefix.

Let me give another approach to think about this. Might be a little bit to much theory, but just for education and reasoning. Note that the famous Dyck language is defined as
$$
 D := \{ w \in \{a,b\}^{\ast} :  |u|_a \le |u|_b \mbox{for each prefix $u$ of $w$ and } |w|_a = |w|_b \}.
$$
It is quite simple to give a grammar for this language, for example
$$
 S \to \varepsilon \mid SS \mid aSb.
$$
Now if we denote by $\mbox{pref}(L)$ the set of prefixes of some language $L$ then your initial language of words having at most as many $a$'s as $b$'s in any prefix is $\mbox{pref}(D)$. And it is quite easy to derive a grammar from it, just make sure every production could be finished any time on the right (for a general argument see this post), hence we could give the grammar
$$
 S \to \varepsilon \mid SS \mid aSb \mid aS \mid a
$$ 
where the production $S \to aS$ could also be leaved out. Now simply note that words fulfilling your restriction have to start with an $a$ and proceed as above.