Just to tie this one up, the cited result in Borceux (Prop 5.6.1) says the following:
Proposition: Let $\mathcal C$ be a finitely-complete category and let $L \mathcal C$ be a reflective subcategory. Let $\mathcal W$ be the class of morphisms inverted by the reflector $r: \mathcal C \to L\mathcal C$. Then $r$ is left exact if and only if $\mathcal W$ is stable under base change.
This continues to hold in the $\infty$-categorical context. Here is a proof adapted from Borceux which works $\infty$-categorically.
Proof: The "only if" direction is clear; we prove "if". First, since $i$ preserves terminal objects we have that the terminal object of $L\mathcal C$ is the terminal object of $\mathcal C$ and in particular is $\mathcal W$-local, so $L$ preserves terminal objects.
So it will suffice to show that $L$ preserves pullbacks. To this end, it is sufficient to show that if we have a natural transformation from the pullback square on the left to the one on the right below, and if the components $B \xrightarrow \sim B'$, $C \xrightarrow \sim C'$, and $D \xrightarrow \sim D'$ are all in $\mathcal W$, then so is the map $A \to A'$.
$\require{AMScd} \begin{CD} A @>>> B\\ @VVV @VVV\\ C @>>> D \end{CD} \qquad \Rightarrow \qquad\begin{CD} A' @>>> B'\\ @VVV @VVV\\ C' @>>> D' \end{CD}$
By pullback-stability and 2/3, the map $B \times_{D'} C \xrightarrow \sim A'$ is in $\mathcal W$. So by 2/3 it will suffice to show that the map $A \to B \times_{D'} C$ is in $\mathcal W$. By pullback-stability and 2/3, the map $D \xrightarrow \sim D \times_{D'} D$ is in $\mathcal W$, so it will suffice by pullback-stability to show that the following two squares are pullbacks:
$\begin{CD} A @>>> B @>>> D \\ @VVV @VVV @VV^\sim V \\
B \times_{D'} C @>>> B \times_{D'} D @>>> D \times_{D'} D \end{CD}$
This can be seen using the following two diagrams:
$\begin{CD} A @>>> B \\ @VVV @VVV \\
B \times_{D'} C @>>> B \times_{D'} D @>>> B \\ @VVV @VVV @VVV \\
C @>>> D @>>> D' \end{CD}
\quad \begin{CD} B @>>> D \\ @VVV @VVV \\
B \times_{D'} D @>>> D \times_{D'} D @>>> D \\ @VVV @VVV @VVV \\
B @>>> D @>>> D' \end{CD}
$
In each case we argue that the bottom-right square and the composite of the lower two squares is a pullback, so the bottom-left square is a pullback. Since the composite of the left two squares is also a pullback, it results that the top-left square is a pullback.