The filters that you're looking for don't exist. Every filter with your property, other than the Frechet filter, maps to an ultrafilter via a finite-to-one map. Let's call the filters with your property Szewczak-Tsaban filters, or S-T filters for short.
Theorem: If $\mathcal F$ is an S-T filter other than the Frechet filter, then there is a finite-to-one map that sends $\mathcal F$ to an ultrafilter.
Lemma: Suppose $\mathcal F$ is an S-T filter and $B \in \mathcal F$. If $\mathcal F$ is not an ultrafilter, then $B-1 \in \mathcal F$.
Proof: Suppose $B \in \mathcal F$ and $B-1 \notin \mathcal F$, and suppose $\mathcal F$ is not an ultrafilter. We will show that $\mathcal F$ is not S-T. To do this, notice that $B \setminus (B-1) \in \mathcal F^+$. (In plain English, $B \setminus (B-1)$ is the set of right-hand endpoints of maximal subintervals of $B$). Using that $\mathcal F$ is not an ultrafilter, we can find some $A_0 \subseteq B \setminus (B-1)$ with $A_0 \in \mathcal F^+$ but $A_0 \notin \mathcal F$. Let $$A = A_0 \cup (A_0+1).$$ Now look at the definition of an S-T filter using these particular values of $A$ and $B$. The set $C$ that we get is precisely equal to $A_0$, which is not in $\mathcal F$ by design. Thus $\mathcal F$ is not S-T. QED(lemma).
Proof of theorem: Fix a non-principal filter $\mathcal F$ other than the Frechet filter, and assume that $\mathcal F$ does not map to an ultrafilter by a finite-to-one map. We'll show $\mathcal F$ is not S-T. Because $\mathcal F$ is not Frechet, there is some infinite $D \subseteq \omega$ such that $\omega \setminus D \in \mathcal F$.
Partition $\omega$ into intervals as follows: if $n \in D$, then $[n,n]$ is in our partition, and we partition $\omega \setminus D$ into maximal intervals (which are all finite because $D$ is infinite). Let $\{I_n : n \in \omega\}$ be an enumeration of this partition. Let $E_0$ be the set of $n$ such that $I_n$ is a maximal interval of $\omega \setminus D$. Notice that $\bigcup_{n \in E_0}I_n = \omega \setminus D \in \mathcal F$.
Consider the finite-to-one map induced by this partition (i.e., the map that sends each element of $I_n$ to $n$). By assumption, this map does not send $\mathcal F$ to an ultrafilter. Therefore there is some $E \subseteq E_0$ such that $\bigcup_{n \in E_0}I_n$ is in $\mathcal F^+$ but not $\mathcal F$.
Let $A = \bigcup_{n \in E}I_n$ and let $$B = \left(\omega \setminus D\right) \cap \left(\left(\omega \setminus D\right)-1\right).$$ By our choice of $E$, we have $A \in \mathcal F^+$. By our lemma and our choice of $D$, we have $B \in \mathcal F$. Now plug this $A$ and $B$ into your definition of an S-T filter. You'll find that what pops out is precisely the set $A$. But by design, we have $A \notin \mathcal F$, so $\mathcal F$ is not S-T. QED