No, this cannot be done in general. Indeed, let $A:=\mathcal A$. You want to express 
$$P(A)=Ef(X,Y)$$
as 
$$\sum_{k=1}^K c_k F(a_k,b_k)=Eg(X,Y),$$
where 
$$f(x,y):=1(x\le2,x-y\le3)$$
and 
$$g(x,y):=\sum_{k=1}^K c_k 1(x\le a_k,y\le b_k).$$
However, for any choice of the numbers $a_k,b_k,c_k$, there will be some $(x_*,y_*)\in\mathbb R^2$ such that such that $f(x_*,y_*)\ne g(x_*,y_*)$. Letting the random pair $(X,Y)$ take value $(x_*,y_*)$ with probability $1$, we get 
$$P(A)=Ef(X,Y)=f(x_*,y_*)\ne g(x_*,y_*)=Eg(X,Y)=\sum_{k=1}^K c_k F(a_k,b_k),$$ 
so that $P(A)\ne\sum_{k=1}^K c_k F(a_k,b_k)$.