The conjecture is true and holds, in a generalized version, whenever $K$ is an abelian regular subgroup. In fact by Theorem 1.9(d) in O. Tamaschke, Zur Theorie der Permutationsgruppen mit regulärer Untergruppe. I, Math. Zeit. (1963) 80 328–354, there is an even more general result that holds when $K$ is not necessarily abelian. A special case of this theorem is that $\langle \mathbb{1}_K, \phi_1, \ldots, \phi_\ell \rangle_\mathbb{C}$ has the structure of a unitary central $S$-ring. In particular, it is closed under the natural product.
Below I'll give a proof in the cyclic case, using some ideas from Wolfgang Knapp On Burnside's Method, J. Alg. (1995) 175 644–660. In outline, what we show is that $\langle \mathbb{1}_K, \phi_1, \ldots, \phi_\ell \rangle$ under multiplication is isomorphic to the Schur ring for $G$ with respect to the product that sends two $H$-invariant subsets to their intersection.
Setup. Let $\zeta \in \mathbb{C}$ be a primitive $d$-th root of unity. For $j \in \{0,\ldots, d-1\}$ let
$$v_j = \sum_{i=0}^{d-1} \zeta^{-ij} k^i \in \mathbb{C}K.$$
The $v_j/d$ are the orthogonal primitive idempotents in the commutative group algebra $\mathbb{C}K$. Since $v_j k = \zeta^j v_j$, the subspace $\langle v_j \rangle$ affords the character $\theta^j$. Let $\odot$ denote the product on $\mathbb{C}K$ defined by multiplying coefficients of each $k^j$: thus if $d=3$ then
$$(a_0+a_1k+a_2k^2) \odot(b_0+b_1k+b_2k^2) = a_0b_0 + a_1b_1 k + a_2b_2k^2.$$
Let $\mathcal{C}(K)$ denote the character ring of $K$. Consider the linear map $\mathcal{C}(K) \rightarrow \mathbb{C}K$
$$\chi \mapsto \sum_{i=0}^{d-1} \overline{\chi(k^j)}k^j.$$
Since the image of $\theta^j$ is $v_j$, this map is a linear isomorphism. Observe that
$$v_j \odot v_{j'} = v_{j + j' \,\mathrm{mod}\; d}.$$
Therefore $\mathcal{C}(K)$, with its usual product, corresponding to the tensor product of modules, is isomorphic to $(\mathbb{C}K, \odot)$.
Proof of conjecture. Let $G$ act on $\{0,1,\ldots,d-1\}$. Let $H \le G$ be the point stabiliser of $0$. Let $\mathcal{O}_0, \mathcal{O}_1, \ldots, \mathcal{O}_\ell$ be the orbits of $H$ on $\{0,1,\ldots,d-1\}$, ordered so that $\mathcal{O}_0 = \{0\}$. (Note that the number of orbits is the rank of the $G$-action, which is the number of irreducible constituents of $\pi$, namely $\ell+1$.) Identify the $G$-set $\{0,1,\ldots,d-1\}$ with the regular subgroup $K$ by $j \mapsto k^j$. The orbit sums are then
$$e_m = \sum_{j \in \mathcal{O}_m} k^j \in \mathbb{C}K$$
for $m \in \{0,1,\ldots,\ell\}$. By definition $S = \langle e_0, e_1, \ldots, e_\ell \rangle_\mathbb{C}$ is the Schur ring for $G$. The $e_m$ are Schur's `primitive elements' and are primitive idempotents for the $\odot$ product. (The Schur ring is also a unital ring under the normal product on $\mathbb{C}K$, but, perhaps surprisingly, we do not need this fact here.)
By our identification, $\mathbb{C}K$ affords the natural permutation module for $G$. Let
$$\mathbb{C}K = V_0 \oplus V_1 \oplus \cdots \oplus V_\ell$$
be its direct sum decomposition into irreducible representations, where
$$V_0 = \langle 1_K + k + \cdots + k^{d-1} \rangle$$
affords the trivial representation, and $V_i$ affords the character $\pi_i$ in the question. Note that $V_0 = \langle v_0 \rangle$. More generally, let $B_i$ be the set of $j$ such that $\langle \pi_i\!\!\downarrow_K, \theta^j \rangle = 1$. Then $V_i = \langle v_j : j \in B_i\rangle$. Moreover, by Frobenius reciprocity
$$\langle \pi_i \!\!\downarrow_H, \mathbb{1}_H \rangle = \langle \pi_i, \mathbb{1}_H\!\!\uparrow^G \rangle = \langle \pi_i, \pi \rangle = 1$$
for each $i$, so each $V_i$ has a unique (up to scalars) $H$-invariant vector. Since $e_0 = \frac{1}{d}(v_0 + v_1 + \cdots + v_{d-1})$ is $H$-invariant, taking the projection into $V_i$ we see that a suitable choice for this $H$-invariant vector is $\sum_{j \in B_i} v_j$. Therefore an alternative basis for the Schur ring $S$ is
$$\big\{ \sum_{j \in B_i} v_j : j \in \{0,1,\ldots, d-1\bigr\}.$$
Under the isomorphism in the 'setup', we have
$$\phi_i = \sum_{j \in B_i} \theta^j \mapsto \sum_{j \in B_i} v_j.$$
Therefore, by the inverse of this isomorphism, $(S, \odot)$ is isomorphic to $\langle \mathbb{1}_K, \theta_1, \ldots, \theta_\ell \rangle$. In particular, the linear span of characters is closed under the product in $\mathcal{C}(K)$. $\Box$
