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No, this cannot be done in general. Indeed, let $A:=\mathcal A$. You want to express $$P((X,Y)\in A)=Ef(X,Y)$$ 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 $f(x_*,y_*)\ne g(x_*,y_*)$. Letting the random pair $(X,Y)$ take value $(x_*,y_*)$ with probability $1$, we get $$P((X,Y)\in A)=Ef(X,Y)=f(x_*,y_*)\ne g(x_*,y_*)=Eg(X,Y)=\sum_{k=1}^K c_k F(a_k,b_k),$$ so that $P(A)\ne\sum_{k=1}^K c_k F(a_k,b_k)$.