We seem to be talking past each other in the limited space provided by the comments, so maybe I can express myself better in the room provided by the answer box. You indicated that you were focused on countable discrete groups. For countable discrete $G$ and $H < G$ the following are equivalent.

There exists a constant $c > 0$ such that for any measure-preserving ergodic action of $G$ on a standard probability space $(X,\mu)$ there is a measurable $H$-invariant subset $A \subseteq X$ on which $H$ acts ergodically and with $\mu(A) \geq c$.

$H$ has finite index in $G$.

(not 2) $\Rightarrow$ (not 1). If $H$ has infinite index, consider the ergodic action of $G$ by shifts on $[0,1]^{G/H}$ with the usual product measure. $H$ doesn't act ergodically on any set of positive measure, as the components of its ergodic decomposition are null. More concretely, if $A \subseteq [0,1]^{G/H}$ is an $H$-invariant set of positive measure, there are disjoint sets $B,C \subseteq [0,1]$ such that a positive measure of elements of $A$ send the coset $H$ to something in $B$, and the same for $C$. Since the $H$ action doesn't shift this coset, this shows $H$ doesn't act ergodically on any set of positive measure. (A better way of writing this argument is simply that $f \mapsto f(H)$ is a null-to-one $H$-invariant Borel function from $[0,1]^{G/H}$ to $[0,1]$, which is enough to preclude ergodicity of the $H$ action on any non-null set.) Thanks to Robin for fixing the error in the original answer.

2 $\Rightarrow$ 1. Say $H$ has index $n$ in $G$, and fix coset representatives $g_1, \ldots, g_n$. Fix a measure-preserving ergodic action of $G$ on $(X, \mu)$. Suppose that $B \subseteq X$ has positive measure and is $H$-invariant. Then $\bigcup_{i \leq n} (g_i \cdot B)$ is $G$-invariant and $\mu$-positive, so by ergodicity has measure $1$. This implies that $\mu(B) \geq 1/n$. So picking $A$ to be an $H$-invariant set of smallest positive measure, $H$ acts ergodically on $A$ and $\mu(A) \geq 1/n$. Thus $c = 1/n$ works for all actions.