I wasn't able to solve the problem based only on Lopsy's hint, so here's a bit more.
First, it's all well and good to apply nifty reflection tricks, but they're a lot easier to find if you already know the result you're aiming for; so let's first mechanically derive the result using generating functions and then think about how to get it more elegantly.
The sequences that don't fulfill the requirement consist of a segment (possibly empty) in which $B$ is in the lead, followed by a segment (possibly empty) in which $A$ is in the lead. Such segments where the lead doesn't change are counted by the Catalan numbers, so these invalid sequences are counted by a convolution of the Catalan numbers with themselves (with the sum running over the point where the lead changes). In terms of generating functions, that means that the generating function $G$ of the invalid sequences is the square of the generating function $C$ of the Catalan numbers. With
$C(x)=\frac{1-\sqrt{1-4x}}{2x}\;,$
that yields
$G(x)=C(x)^2=\frac1x\left(\frac{1-\sqrt{1-4x}}x-1\right)=\frac{C(x)-1}x\;.$
Thus, $G$ is just $C$ with the constant term removed and shifted down by one, that is, $G_n=C_{n+1}$.
Knowing the result, it's a bit easier to see how to apply reflection. The problem in pursuing Lopsy's hint is that it's not obvious how to get a bijection – it's easy to reflect the part below the diagonal upward, but it's not clear what bijection that establishes. Knowing that we want to end up with the Catalan numbers one higher, we can use the extra slot to make the reflected sequence unique: By inserting an up-step before the reflected segment and a down-step after it, we get a bijection from the invalid sequences to the diagonal-avoiding sequences with two more steps, since the turning point is now uniquely marked as the last intersection with the diagonal in the new sequence.