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…

======================

**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.