Let $\mathcal{P}(\omega)/fin$ denote the Boolean algebra formed from $\mathcal{P}(\omega)$ by modding out by the ideal $fin$ of finite subsets of $\omega$. As a first pass at the intended question, consider the following:

Question 0: Are there two filters $F$ and $G$ in $\mathcal{P}(\omega)/fin$ such that there is a unique ultrafilter extending $F \cup G$?

The answer, of course, is yes: consider the case where $F$ is already an ultrafilter, and $G$ is some filter such that $G \subseteq F$. We might therefore ask (what seems to be) a harder question. Given an ideal $I$ in $\mathcal{P}(\omega)/fin$, notice that

$$\{a \in \mathcal{P}(\omega)/fin \: : \: a \geq I\}$$

(where $a \geq I$ iff $(\forall b \in I)[a \geq b]$) is a filter; call such filters regular filters (I made up this terminology, and I would be glad to know if there is already a word for such objects). Now we can ask:

Question 1: Are there two regular filters $F$ and $G$ such that there is a unique ultrafilter extending $F \cup G$?

Via Stone duality, this question can be rephrased (I believe) in topological terms:

Question 1$'$: Are there two regular closed subsets $C,D \subset \omega^{\*}$ such that $C \cap D$ is a singleton?

Here, a regular closed set is simply a set which is equal to the closure of its interior, and $\omega^{\*} = \beta \omega \setminus \omega$, the space of all non-principal ultrafilters on $\omega$ (i.e. the Stone space of $\mathcal{P}(\omega)/fin$). If $C$ and $D$ are witnesses to a positive answer to question 1$'$, then $int(C)$ and $int(D)$ must be disjoint, in which case the ideals corresponding to these open sets form a gap; this is the basis for my original interest in this question.