First of all here, your $D_{\alpha,\beta}(f||g)$ is not properly defined. It is not a function of $f$ and $g$ (for given $\alpha$ and $\beta$), as it also depends on the joint distribution of $X$ and $Y$. Also, you need to ensure that $P(Y>\beta)>0$ and $P(Y\le\beta)>0$ ($\beta$ being in the support of $Y$ would not be enough). 

Further, at least one of the probabilities $P(\beta<X\le b)$ or $P(b<X\le\beta)$ equals $0$, depending on whether $\beta>b$ or $\beta\le b$. So, at least one of the conditional probabilities $P(X\le b|X>\beta)$ or $P(X>b|X\le\beta)$ equals $0$, provided that both are defined. So, if $0<\alpha<1$ and $Y=X$, then the right-hand side of (1) never equals $0$, which violates the second one of the two conditions of being a divergence that you referred to. 

If now $\alpha=1$ and $Y=X$, then the right-hand side of (1) equals $0$ only if $P(X\le b)=1$ and $P(X>\beta)>0$. Similarly, if $\alpha=0$ and $Y=X$, then the right-hand side of (1) equals $0$ only if $P(X>b)=1$ and $P(X\le\beta)>0$. 

To conclude, in no way is $D_{\alpha,\beta}(f||g)$ a divergence.