This does not hold for all commutative rings, but it does hold for Noetherian rings and for valuation rings (assuming the convention that filters don't contain $\emptyset,$ or else $\mathcal{P}(R)$ is a trivial counterexample for any ring).
Suppose $R$ is a commutative ring, $F$ is an ideal-like filter on $R,$ and $U_0 \in F$ is such that there is no ideal $I \in F \cap \mathcal{P}(U_0).$ Recursively choose $U_n \in F$ such that $U_{n+1}+U_{n+1} \subset U_n$ and $U_{n+1}=U_{n+1}R.$ Since no $U_n$ is empty or an ideal, we immediately have $U_{n+1} \subsetneq U_n.$
For $n \ge 1,$ we recursively choose $u_n \in U_n$ for $n \ge 1$ as follows: first, let $u_1$ be any element of $U_1 \setminus U_2.$ Suppose $u_1, \ldots, u_n$ have been chosen. An inductive argument shows that $I_n:=\langle u_1, \ldots, u_n \rangle \subset U_0,$ so $I_n \not \in F.$ Let $u_{n+1}$ be any element of $U_{n+1} \setminus I_n.$ Then $\{I_n\}$ is a strictly increasing sequence of ideals, so $R$ is non-Noetherian. Furthermore, $\langle u_1 \rangle$ and $\langle u_2 \rangle$ are incomparable, so $R$ is not a valuation ring.
Now we'll construct a commutative ring $R$ with an ideal-like filter $F$ not generated by ring ideals. Let $R = C^{\infty}(S^1)$ be the ring of smooth real-valued functions on the circle. Every maximal ideal in $R$ is of the form $\mathfrak{m}_x=\{f \in R: f(x)=0\}$ for some $x \in S^1$ (see https://math.stackexchange.com/a/183213/210610). Let
$A_n=\{f \in R: \lambda(f^{-1}(0))> 1- \frac{1}{n}\},$ and let $F$ be the filter on $R$ generated by the $A_n$'s, i.e.
$$F=\bigcup_{n=1}^{\infty} \{X \subset R: A_n \subset X\}.$$
It's clear that every $X \in F$ is infinite (in fact, $|X|=|\mathbb{R}|$) and that for every finite $S \subset R \setminus \{0\},$ $R \setminus S \in F.$ We will show that there is no proper ideal in $F.$ Suppose $I \in F$ is a proper ideal on $R.$ There is $x$ such that $I \subset \mathfrak{m}_x \in F.$ But for every $n,$ there is a bump function in $A_n \setminus \mathfrak{m}_x,$ contradiction. Thus, $U=R \setminus \{1\} \in F$ does not contain any ideal in $F.$
Finally, we verify that $F$ is ideal-like. Fix $U \in F.$ Let $n$ be such that $A_n \subset U.$ Then $V=A_{2n}$ satisfies $V=VR$ and
$V+V \subset A_n \subset U.$
Original answer:
This does not hold for all commutative rings, but if $R$ is a Dedekind domain, and we include the filter convention that $\emptyset \not \in F,$ then an ideal-like filter $F$ is generated by ring ideals.
Suppose $F$ is a counterexample. Let $U \in F$ be such that there is no ideal $I \in F \cap \mathcal{P}(U).$ Let $V, W \in F$ be such that $V+V+V \subset U$ and $WR \subset V.$ We may assume $W=WR.$ Since $W$ is nonempty and not an ideal, we can fix a nonzero non-unit $w \in W,$ and prime ideals $P_i$ such that $\langle w \rangle = \prod_{i=1}^n P_i.$
Let $A \subset \{1,\ldots, n\}$ be maximal such that $\prod_{i \in A} P_i \in F.$ Let $X \in F$ be such that $X=XR$ and $\sum_{i=1}^n X \subset W.$ For $j \in A,$ set $x_j=0,$ and otherwise let $x_j$ be an arbitrary element of $X \cap \prod_{i \in A} P_i \setminus \prod_{i \in A \cup \{j\}} P_i,$ which is nonempty since $X \cap \prod_{i \in A} P_i \in F.$ Let $x, y$ be such that $\langle x, y \rangle = \langle x_i: i \le n \rangle \subset W.$ Consider
$$I:=\langle w, x, y \rangle \subset (W + W + W) \cap \prod_{i \in A} P_i \subset U \cap \prod_{i \in A} P_i.$$ Since $\prod_{i=1}^n P_i \subset I
\subsetneq \prod_{i \in A} P_i \in F,$ there is $B \supsetneq A$ such that $I = \prod_{i \in B} P_i.$
Fix $j \in B \setminus A.$ Then $A \cup \{j\} \subset B$ and $x_j \in \langle x, y \rangle,$ so $x_j \in \prod_{i \in B} P_i \setminus \prod_{i \in A \cup \{j\}} P_i = \emptyset,$ contradiction.
