Sadly, the example that you're looking for does not exist. 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, then there is a finite-to-one map that sends $\mathcal F$ either to the Frechet filter or to an ultrafilter.

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 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 $A$ 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 $\mathcal F$ that does not map to either the Frechet filter or an ultrafilter by a finite-to-one map. We'll show this filter isn't S-T. Let $I_0,I_1,\dots,I_n,\dots$ be a partition of $\omega$ into intervals such that, for some $D \subseteq \omega$ with $\omega \setminus D$ infinite, we have $\bigcup_{n \in D}I_n \in \mathcal F$ (this is possible because $\mathcal F$ is not sent to the Frechet filter by a finite-to-one map; see, e.g., Proposition 9.4 in Blass's handbook article).

Let $E_0 = D \setminus (D-1)$ (which is infinite because $\omega \setminus D$ is infinite). If either $\bigcup_{n \in E}I_n$ or $\bigcup_{n \notin E}I_n$ is in $\mathcal F$ for every $E \subseteq E_0$, then $\mathcal F$ gets sent to an ultrafilter under the finite-to-one map $I_n \mapsto n$. So fix some $E \subseteq E_0$ with neither $\bigcup_{n \in E}I_n$ nor $\bigcup_{n \notin E}I_n$ in $\mathcal F$. In other words, we have $E \subseteq E_0$ such that $\bigcup_{n \in E}I_n$ is in $\mathcal F^+$ but not in $\mathcal F$.

Let $A = \bigcup_{n \in E}I_n$ and let $$B = \left(\bigcup_{n \in D}I_n\right) \cap \left(\left(\bigcup_{n \in D}I_n\right)-1\right).$$ 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

I think you should actually be able to go beyond this theorem and prove something stronger:

Guess: $\mathcal F$ is an S-T filter if and only if one of the following three things holds:

(1) $\mathcal F$ is the Frechet filter.

(2) $\mathcal F$ is an ultrafilter.

(3) $\mathcal F$ is the Stone dual of a minimal left ideal of $(\omega^*,+)$.

Every filter satisfying $(3)$ maps to an ultrafilter by a finite-to-one map, so this is strictly stronger than what we've proved. I worked on this quite a bit this afternoon, but I can't quite seem to make it work. I may think about it some more, and if I make any more progress then I'll be sure to let you know.

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.

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