The answer seems to be yes.
Well-order $V$ and collect $D$ inductively. Let $D_v$ be the set collected after considering $v$; for the smallest $v$ set $D_v=\{v\}$.
When considering some further $v$, check whether it is covered by $$ \bigcup\{e\in E\colon e\cap D_u\neq\varnothing \text{ for some $u<v$}\}. $$ If yes, set $D_v=\bigcup _{u<v} D_u$. Otherwise, set $D_v= \bigcup _{u<v} D_u \cup\{v\}$. Notice that in the latter case no edge containing $v$ contains any other vertex in $D_v$.
Finally, $D=\bigcup_v D_v$ is what you want. A vertex $v\in D$ cannot be removed, as it is contained in no edge containing any other $u\in D$.
The condition on finite degrees is not used…
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This is a previous (wrong) answer, as I misread the definition of a dominating set.
Choose $V=\{1,2,\dots\}$ and let the edges be $e_i=\{i,i+1,\dots\}$. The dominating sets are precisely infinite ones.