It is a classical fact (e.g. here) that for an (accessible) right adjoint comonad on a (sheaf) topos, its Eilenberg-Moore category of coalgebras is itself a (sheaf) topos.
I suppose this remains true for $\infty$-toposes, for hypercomplete $\infty$-stack $\infty$-toposes at least? (But not assuming that the comonad is idempotent.)
If $E$ is an ∞-topos and $T: E \to E$ is an accessible left exact comonad, then indeed the ∞-category $E^T$ of $T$-coalgebras is an ∞-topos. Moreover, it is hypercomplete if $E$ is. I will first show:
Lemma. Let $E'$ be presentable and let $U:E'\to E$ be a conservative functor that preserves colimits and pullbacks. Then $E'$ is an ∞-topos.
Proof. Recall that a presentable ∞-category $E'$ is an ∞-topos iff it satisfies descent, which means the following: If $a,b: K^\triangleright \to E'$ are cones, $\phi: a \Rightarrow b$ is a natural transformation, $b$ is a colimit, and $\phi$ is cartesian on $K$, then $a$ is a colimit iff $\phi$ is cartesian. In our situation, by the hypotheses on $U$, $a$ is a colimit iff $Ua$ is a colimit iff $U\phi$ is cartesian iff $\phi$ is cartesian. //
If $T$ is an accessible comonad on $E$, $E^T$ is accessible since it can be written as a limit of accessible ∞-categories. Moreover, the forgetful functor $U: E^T\to E$ creates colimits and is conservative ([1] 3.2.2.5 and 3.2.2.6). If $T$ is left exact, then $U$ is also left exact ([2] Corollary 5.5).
So the lemma implies that $E^T$ is an ∞-topos. Moreover, the forgetful functor $U$ is the left adjoint of a geometric morphism. Since $U$ preserves connectivity and is conservative, $E'$ is hypercomplete if $E$ is.
[1] J. Lurie, Higher Algebra
[2] E. Riehl and D. Verity, Completeness results for quasi-categories of algebras, homotopy limits, and related general constructions
