$\newcommand\LL{\mathcal L}$**Brief answer:** The Scott topology is not Hausdorff, and therefore we have to deal here with the set of limits, rather than with the (unique) limit. Here, for the set $\LL_D$ of limits of the identity net on $D$ we have 
\begin{equation*}
	\LL_D=L_s:=\{y\in X\colon x\le s\}, 
\end{equation*}
where 
\begin{equation*}
	s:=s_D:=\sup D=\bigvee D. 
\end{equation*}

**Details:** 

First, a review of definitions. 

Let $X$ be a set with a partial order $\le$. 

 - A point $b\in X$ is an *upper bound* of a subset $A$ of $X$ if $x\le b$ for all $x\in A$. 
 
 - A nonempty subset $D$ of $X$ is *directed* if any finite subset of $D$ has an upper bound in $D$. 
 
 - A point $s\in X$ is the *supremum* of a subset $A$ of $X$ and denoted by $\sup A$ or $\bigvee A$ if $s$ is the least upper bound of $A$, that is, if $s$ is an upper bound of $A$ and for any upper bound $b$ of $A$ we have $s\le b$. 
 
 - A subset $A$ of $X$ is an *upper set* if for any $x\in A$ we have $U_x\subseteq A$, where $U_x:=\{y\in X\colon x\le y\}$. 
  
 - A subset $A$ of $X$ is a *lower set* if for any $x\in A$ we have $L_x\subseteq A$, where $L_x:=\{y\in X\colon y\le x\}$. 
 
 - The pair $(X,\le)$ of a set $X$ with a partial order $\le$ is a *directed-complete partial order (dcpo)* if each directed subset $D$ of $X$ has a supremum. In what follows, $(X,\le)$ will be a dcpo. 
 
 - A subset $G$ of $X$ is *open in the Scott topology*, or *Scott-open*, if $G$ is an upper set and for any directed subset $D$ of $X$ such that $\sup D\in G$ we have $D\cap G\ne\emptyset$. Equivalently, a subset $F$ of $X$ is *closed in the Scott topology*, or *Scott-closed*, if $F$ is a lower set and for any directed subset $D$ of $F$ we have $\sup D\in F$. 
 
 -- The identity net $I_D$ on a directed subset $D$ of $X$ is the map $D\ni x\mapsto I_D(x):=x\in X$. 
 
 -- The set $\LL_D$ of limits of the identity net $I_D$ is defined by the following condition on $x\in X$:
 
\begin{equation*}
\begin{aligned} 
x\in\LL_D\iff &\text{for any Scott-open $G$ such that $x\in G$ } \\ 
&\text{there is some $z_G\in D$ such that $U_{z_G}\subseteq G$.}
\end{aligned} 	
\end{equation*}
 
 
---

Let now $D$ be any directed subset of $X$. 

Suppose that $x\in\LL_D$. 
We want to show that then $x\in L_s$. Suppose the contrary. Then $x\in V:=X\setminus L_s$, and the set $V$ is Scott-open (since the set $L_x$ is Scott-closed for any $x\in X$). On the other hand, $z_V\in D$ and $s=\sup D$, so that $z_V\le s$, that is, $s\in U_{z_V}$. Also, by the definition of $\LL_D$, we have $U_{z_V}\subseteq V$. So, $s\in U_{z_V} \subseteq V=X\setminus L_s$, which contradicts the obvious fact that $s\in L_s$. 

We see that $x\in L_s$ for any $x\in\LL_D$, that is, 
\begin{equation*}
	\LL_D\subseteq L_s. \tag{1}\label{1}
\end{equation*}

Let us now show that $s\in\LL_D$. Take any any Scott-open $G$ such that $s\in G$. Since  $G$ is Scott-open and $s=\sup D\in G$, we have $D\cap G\ne\emptyset$. Take now any $z_G\in D\cap G$. Since $z_G\in G$ and the Scott-open set $G$ is an upper set, we get $U_{z_G}\subseteq G$. Since $z_G\in D$, we see that indeed $s\in\LL_D$. 

Note also that for any $x\in\LL_D$ we have $L_x\subseteq\LL_D$. Therefore and because $s\in\LL_D$, we see that $L_s\subseteq\LL_D$. 

Now, in view of \eqref{1}, we conclude that 
\begin{equation*}
	\LL_D=L_s=L_{\sup D},
\end{equation*}
as claimed.