This is more like a long comment. Let us assume $\alpha\in[0,1)$ (the case of $\alpha\in(-1,0]$ is similar). As $\sin B>0$ for $B\in (0,\pi)$,  the inequality $f(A,B)<0$ amounts to 
$$
\alpha\sin A<\sin B-\sin(A+B),\quad -[\sin B+\sin(A+B)]<\alpha\sin A.\quad (\star)
$$
Notice that 
$\sin A=2\sin\left(\frac{A}{2}\right)\cos\left(\frac{A}{2}\right)$, 
$\sin B-\sin(A+B)=-2\sin\left(\frac{A}{2}\right)\cos\left(\frac{A}{2}+B\right)$ and 
$\sin B+\sin(A+B)=2\cos\left(\frac{A}{2}\right)\sin\left(\frac{A}{2}+B\right)$. 
Substituting in $(\star)$ and cancelling the positive terms 
$2\sin\left(\frac{A}{2}\right)$ and $2\cos\left(\frac{A}{2}\right)$, we obtain the equivalent inequalities 
$$
\alpha\cos\left(\frac{A}{2}\right)<-\cos\left(\frac{A}{2}+B\right), \quad 
-\sin\left(\frac{A}{2}+B\right)<\alpha\sin\left(\frac{A}{2}\right).\quad (\star\star)
$$
In $(\star\star)$, the LHS of the first inequality and the RHS of the second are non-negative. Hence $\frac{A}{2}+B$  - which belongs to $\left(0,\frac{3\pi}{2}\right)$ - must be in the second or the third quadrant; otherwise, the first inequality in $(\star\star)$ does not hold. Let us analyze these cases separately:

 - If $\frac{\pi}{2}\leq\frac{A}{2}+B\leq\pi$, then the second inequality in $(\star\star)$ holds automatically (its RHS is always non-negative); and the first one can be written as 
$$\alpha\cos\left(\frac{A}{2}\right)<\cos\left(\pi-\frac{A}{2}-B\right).$$ Applying the strictly decreasing function $\cos^{-1}:[0,1]\rightarrow\left[0,\frac{\pi}{2}\right]$ yields: 
$$\pi-\frac{A}{2}-B<\cos^{-1}\left(\alpha\cos\left(\frac{A}{2}\right)\right).$$
This of course implies $\frac{A}{2}+B\geq\frac{\pi}{2}$. But we also need $\frac{A}{2}+B\leq\pi$. Combining these, the bounds for $B$ in terms of $A\in(0,\pi)$ are given by
$$
\pi-\frac{A}{2}-\cos^{-1}\left(\alpha\cos\left(\frac{A}{2}\right)\right)\leq B\leq\pi-\frac{A}{2}.
$$
The difference of the two bounds is $\cos^{-1}\left(\alpha\cos\left(\frac{A}{2}\right)\right)$. Consequently, the contribution to the area of 
$\{(A,B)\mid f(A,B)<0\}$ is 
$$
\int_{\{(A,B)\mid f(A,B)<0,\, \frac{\pi}{2}\leq\frac{A}{2}+B\leq\pi\}}\mathbf{1}=
\int_{0}^\pi\cos^{-1}\left(\alpha\cos\left(\frac{A}{2}\right)\right){\rm{d}}A\quad (1).
$$
 - If $\pi\leq\frac{A}{2}+B\leq\frac{3\pi}{2}$, all terms appearing in $(\star\star)$ are non-negative. We first rewrite these inequalities as
$$
\alpha\cos\left(\frac{A}{2}\right)<\cos\left(\frac{A}{2}+B-\pi\right), \quad 
\sin\left(\frac{A}{2}+B-\pi\right)<\alpha\sin\left(\frac{A}{2}\right). 
$$
Next applying strictly monotonic functions 
$\cos^{-1}:[0,1]\rightarrow\left[0,\frac{\pi}{2}\right]$
and $\sin^{-1}:[0,1]\rightarrow\left[0,\frac{\pi}{2}\right]$ to them results in:
$$
\frac{A}{2}+B-\pi<\cos^{-1}\left(\alpha\cos\left(\frac{A}{2}\right)\right),\quad 
\frac{A}{2}+B-\pi<\sin^{-1}\left(\alpha\sin\left(\frac{A}{2}\right)\right).
$$
Hence the upper bound 
$$
B<\pi+\min\left\{\cos^{-1}\left(\alpha\cos\left(\frac{A}{2}\right)\right),\sin^{-1}\left(\alpha\sin\left(\frac{A}{2}\right)\right)\right\}-\frac{A}{2}
$$
which of course implies $\frac{A}{2}+B\leq\frac{3\pi}{2}$. But $\pi\leq\frac{A}{2}+B$ is also required. We therefore arrive at the bounds for $B$ in terms of $A\in(0,\pi)$:
$$
\pi-\frac{A}{2}\leq B\leq 
\pi+\min\left\{\cos^{-1}\left(\alpha\cos\left(\frac{A}{2}\right)\right),\sin^{-1}\left(\alpha\sin\left(\frac{A}{2}\right)\right)\right\}-\frac{A}{2}.
$$  
The difference of the bounds is $\min\left\{\cos^{-1}\left(\alpha\cos\left(\frac{A}{2}\right)\right),\sin^{-1}\left(\alpha\sin\left(\frac{A}{2}\right)\right)\right\}$. Therefore, the contribution to the area of 
$\{(A,B)\mid f(A,B)<0\}$ is 
$$
\int_{\{(A,B)\mid f(A,B)<0,\, \pi\leq\frac{A}{2}+B\leq\frac{3\pi}{2}\}}\mathbf{1}\\
=\int_{0}^\pi\min\left\{\cos^{-1}\left(\alpha\cos\left(\frac{A}{2}\right)\right),\sin^{-1}\left(\alpha\sin\left(\frac{A}{2}\right)\right)\right\}{\rm{d}}A.\quad (2)
$$
Adding $(1)$ and $(2)$, the area of $\{(A,B)\mid f(A,B)<0\}$ turns out to be 
$$
\int_{0}^\pi\cos^{-1}\left(\alpha\cos\left(\frac{A}{2}\right)\right){\rm{d}}A\\+
\int_{0}^\pi\min\left\{\cos^{-1}\left(\alpha\cos\left(\frac{A}{2}\right)\right),\sin^{-1}\left(\alpha\sin\left(\frac{A}{2}\right)\right)\right\}{\rm{d}}A.
$$ 
So the question is if the quantity above coincides with $\frac{\pi^2}{2}$ for all $\alpha\in [0,1)$.