I think there is something intriguing and slightly mysterious going on here.

First, my proposed definition would be slightly different from Dimitri Chikladze's. I agree that the natural generalization of the slice categories $\mathcal V /C$ for $C \in \mathcal V$ should probably be the comodule categories $\mathbf{Comod}(C)$ for $C$ a commutative monoid in $\mathcal V$. But to me, the fundamental way to characerize local cartesian closedness of $\mathcal V$ is that for $f: C \to D$ in $\mathcal V$, the reindexing functor $f_! : \mathcal V/C \to \mathcal V/D$ (defined by postcomposition) not only has a right adjoint $f^\ast$ (defined by pullback), but $f^\ast$ itself has a right adjoint $f_\ast$ (the local hom functor). So I would say:

Definition: A symmetric monoidal category $\mathcal V$ is locally closed if for every commutative monoid homomorphism $f: C \to D$ in $\mathcal V$, the reindexing functor $f_!$ fits into an adjoint string $f_! \dashv f^\ast \dashv f_\ast$ satisfying Beck-Chevalley conditions.

The strange thing here is when we look at examples. If $\mathcal V$ has certain nice limits and colimits (e.g. $\mathcal V = \mathsf{Ab}^{\mathrm{op}}$), then we always have an adjoint string $f^! \dashv f_! \dashv f^\ast$ (where $f^\ast$ and $f^!$ are respectively induction and coinduction of modules if $\mathcal V = \mathsf{Ab}^\mathrm{op}$). But the extra adjoint $f^!$ is on the wrong side! In order for $f^\ast$ to have a right adjoint, it must be flat when viewed as a monoid homomorphism in $\mathcal V^\mathrm{op}$.

I can't think of a noncocartesian example of a $\mathcal V^\mathrm{op}$ for which all commutative monoid homomorphisms are flat -- even $\mathsf{Vect}_k$ for $k$ a field doesn't seem to work. So it looks like in noncartesian cases, we're already doing what we do with cartesian categories like $\mathsf{Cat}$ -- exponentiability of a morphism is an important property that one is often interested in, but it's just not reasonable to expect every morphism to be exponentiable.

But perhaps there are natural subcategories of all commutative comonoids which have better exponentiability properties in some cases.

I think there is something intriguing and slightly mysterious going on here.

First, my proposed definition would be slightly different from Dimitri Chikladze's. I agree that the natural generalization of the slice categories $\mathcal V /C$ for $C \in \mathcal V$ should probably be the comodule categories $\mathbf{Comod}(C)$ for $C$ a commutative comonoid in $\mathcal V$. But to me, the fundamental way to characerize local cartesian closedness of $\mathcal V$ is that for $f: C \to D$ in $\mathcal V$, the reindexing functor $f_! : \mathcal V/C \to \mathcal V/D$ (defined by postcomposition) not only has a right adjoint $f^\ast$ (defined by pullback), but $f^\ast$ itself has a right adjoint $f_\ast$ (the local hom functor). So I would say:

Definition: A symmetric monoidal category $\mathcal V$ is locally closed if for every commutative comonoid homomorphism $f: C \to D$ in $\mathcal V$, the reindexing functor $f_!$ fits into an adjoint string $f_! \dashv f^\ast \dashv f_\ast$ satisfying Beck-Chevalley conditions.

The strange thing here is when we look at examples. If $\mathcal V$ has certain nice limits and colimits (e.g. $\mathcal V = \mathsf{Ab}^{\mathrm{op}}$), then we always have an adjoint string $f^! \dashv f_! \dashv f^\ast$ (where $f^\ast$ and $f^!$ are respectively induction and coinduction of modules if $\mathcal V = \mathsf{Ab}^\mathrm{op}$). But the extra adjoint $f^!$ is on the wrong side! In order for $f^\ast$ to have a right adjoint, it must be flat when viewed as a monoid homomorphism in $\mathcal V^\mathrm{op}$.

I can't think of a noncocartesian example of a $\mathcal V^\mathrm{op}$ for which all commutative monoid homomorphisms are flat -- even $\mathsf{Vect}_k$ for $k$ a field doesn't seem to work. So it looks like in noncartesian cases, we're already doing what we do with cartesian categories like $\mathsf{Cat}$ -- exponentiability of a morphism is an important property that one is often interested in, but it's just not reasonable to expect every morphism to be exponentiable.

But perhaps there are natural subcategories of all commutative comonoids which have better exponentiability properties in some cases.