I'm not entirely sure I understand the notation, because it seems $a\times b$ would be a filter on the set of ordered pairs, not on the set of binary relations (though it would be a filter in the Boolean algebra of binary relations). But, assuming my best guess for what was intended, here's a counterexample. Let both $a$ and $b$ be the filter $\mathcal F$ of cofinite subsets of $\mathbb N$. Let $M$ be the filter on $\mathbb N\times\mathbb N$ generated by all sets of the form $(P\times\mathbb N)\cup(\mathbb N\times Q)$ with $P,Q\in\mathcal F$. Then $M$ is included in both $(\uparrow\mathbb N)\times b$ and $a\times(\uparrow\mathbb N)$. I claim that $M$ is not included in $(\uparrow X)\times(\uparrow Y)$ for any cofinite $X,Y\subseteq\mathbb N$. To see this, consider any such $X$ and $Y$, and let $P\subsetneq X$ and $Q\subsetneq Y$ be slightly smaller but still cofinite sets (for example, remove one element from $X$ and from $Y$). Then $(P\times\mathbb N)\cup(\mathbb N\times Q)$ is in $M$ but not in $(\uparrow X)\times(\uparrow Y)$; i.e., it is not a superset of $X\times Y$. Indeed, if $x\in X-P$ and $y\in Y-Q$ then $(x,y)$ is in $X\times Y$ but not in $(P\times\mathbb N)\cup(\mathbb N\times Q)$. (At the moment, Preview is showing only a part of my answer. I'll post it anyway and hope for the best.)
