I think we only need to assume that almost every $x \in E$ is a point of Lebesgue density of $E,$ which does not require that $E$ is a Lebesgue measurable set. [Indeed, if we can assume that $E$ is a Lebesgue measurable set, then we can conclude much more than what you said, namely that at almost every $x \in E^c$ the Lebesgue density of $E$ at $x$ is equal to $0$ (instead of "is less than $1$"). But I don't see that we need to know anything about the Lebesgue density of $E$ at points not in $E.$]
The result you want follows from the following more precise result.
Theorem: Let $E \subseteq {\mathbb R}^N$, $x \in {\mathbb R}^N$, and assume that $x$ is a point of Lebesgue density of $E.$ Then
$$\lim_{|y| \rightarrow 0}\frac{\delta(x+y)}{|y|} = 0$$
Proof: If not, then $\;\limsup_{|y| \rightarrow 0}\frac{\delta(x+y)}{|y|} > 0,\;$ from which it follows that there exists $\epsilon > 0$ and a sequence $\{y_n\}$ with $y_n \rightarrow x$ such that for each $n$ we have $\delta(x+y_n) \geq \epsilon |y_n|.$ The key idea is to notice that this last inequality tells us there is a sequence of arbitrarily small balls centered at $x,$ each of which contains a "sufficiently big" subball having empty intersection with $E,$ where "sufficiently big" is big enough to contradict the assumption that the Lebesgue density of $E$ at $x$ is equal to $1.$ Specifically, the balls centered at $x$ have radii $|y_n| + \epsilon |y_n|$ and the corresponding subballs are centered at $x+y_n$ and have radii $\epsilon |y_n|.$ Note that the ratio of each subball radius to the corresponding ball radius is ${\epsilon}/({1 + \epsilon}),$ and so the ratio of the volume of each subball to the corresponding volume of the ball is $\left(\frac{\epsilon}{1 + \epsilon}\right)^N.$ Because this ratio of volumes is a positive constant (that depends only on $F,$ $x,$ and $N),$ and no points of $E$ belong to any of the subballs, we cannot have the Lebesgue density of $E$ at $x$ equal to $1,$ which gives a contradiction.
To Summarize: From the assumption "the Lebesgue density of $E$ at $x$ equals $1,$" it follows that, as we shrink down to $x$ using balls centered at $x,$ the ratios of the measures of $E^c$ intersect the balls to the measures of the balls must approach $0$ (because the ratios of the measures of $E$ intersect the balls to the measures of the balls must approach $1$). However, $\delta(x+y_n) \geq \epsilon |y_n|$ provides us with a sequence of balls shrinking down to $x$ in which the ratios of the measures of $E^c$ intersect the balls to the measures of the balls is bounded above $0.$ [Note that the measure of $E^c$ intersect any of the balls is greater than or equal to the measure of the corresponding subball.]
(ADDED NEXT DAY) Something I overlooked: If we don't assume that $E$ is Lebesgue measurable then, in various places above, "measure" and "density" need to be replaced with "outer measure" and "outer density".
Exploit Lebesgue Differentiation Theorem to the locally integrable function $\chi_{E}$:
\begin{align*}
\chi_{E}(x)=\dfrac{1}{m(B(x,r))}\int_{B(x,r)}\chi_{E}(y)dy,~~~~\text{a.e.}~x.
\end{align*}
Best Answer
No, there could be $E^{(1)}$ points outside of $E$. Consider these $\Omega = \mathbb R$ and $E = \mathbb R \setminus \{0\}$. Even though $0 \notin E$, we still have $D(0) = 1$.
If $E$ is open we do have $$ E \subseteq E^{(1)} \subseteq \overline{E}\\ E^c \supseteq E^{(0)} \supseteq \left(\overline{E}\right)^c $$ where I used ${}^c$ for the complement. Any points outside $E^{(0)} \cup E^{(1)}$ are on the boundary of $E$.