$$f(A,B,C,D)=AB(1-C)(1-D)+1$$ is increasing in $A$ and $B$ and decreasing in $C$ and $D$ as well as it satifies $f(A,B,C,D)=1\,\, for\,\, C=1\, or D=1$.
EDIT: Of course there are infinitely many such functions satisfying your conditions. Above is probably one of the simplest one.
One other thing is that one can not define $f(A,B,C,D)\in[0,1]$ satisfying your conditions and still continuous. Here is the reason:
$f$ is decreasing in $C$ and $D$. Let $f$ be somewhere in $[0,1]$ for given $A,B$. As $f\leq1$, $f$ starts decreasing in $C$ and $D$ at most from $1$ and for $C$, $D$, or both very close to $1$ we have $f<1+\Delta$ where $\Delta$ is not arbitrarily close to $0$ when $C\rightarrow 1$ or $D\rightarrow 1$ indicating that $0\leq f(x+\Delta)-f(x)\leq \Delta$ can not be bounded by arbitrary $\Delta$ $\rightarrow$ discontinuity at $C=1$, $D=1$ or $C=1,D=1$