I'm not sure about the suggested equivalence in the last two sentences of your question, but at least the statement about etale group schemes has a negative answer.
That is, it is possible to have an etale group scheme $G rightarrow S$,
with $G$ and $S$ both finite type over a field $k$, but $G$ not finite over $S$.
For example, let $H$ be the constant group scheme ${mathbb Z}/2{mathbb Z}$ over $S$,
let $s$ be some fixed closed point of $S$, and let $G := H setminus 1_s,$
where $1_s$ is the non-zero element of the fibre $({mathbb Z}/2{mathbb Z})_s$.
Then $G$ is open in $H$, hence etale over $S$. Assuming that $S$ is positive dimensional,
it is certainly not finite (we deleted one
point of one fibre), and it is a subgroup scheme of $H$. (If $T$ is an $S$-scheme,
then $G(T)$ is the subgroup of $H(T)$ consisting of points whose values at points of $T$ lying
over $s$ are trivial.)
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