In a left Bousfield localization of the projective model structure on the category of simplicial presheaves, what is the condition that transfinite composition preserves weak equivalences? How about for transfinite composition of fibrant simplicial presheaves?

I apologize if the question doesn't make sense. My question comes from the proof here for Theorem 4.27 showing that the $A^1$ localization functor $L_{A^1}L_{Nis}$ is equivalent to the transfinite composition of $L_{Nis}Sing^{A^1}$, where they show that $L_{Nis}Sing^{A^1}$ takes values in $A^1$-fibrant objects and preserves $A^1$-weak equivalence then conclude the following. So I think the reason for this is that transfinite composition of $A^1$-weak equivalences between fibrant objects are $A^1$-weak equivalence. Is this true and how to prove it?

**Edit**:
The idea of the proof comes as following:
Let $\Phi=L_{A^1}L_{Nis}$, they first show that $\Phi^{\circ \mathbb{N}}X$ is fibrant in $L_{A^1}L_{Nis} sPre(Sm_S)$ for any $X$. Then they show that $\Phi$ preserves $A^1$-local weak equivalences. Since $X\simeq \Phi (X)$ when $X$ is $A^1$-fibrant, applying $\Phi$ tranfinitely to this equivalence should give the following

$\Phi^{\circ \mathbb{N}}(X)\simeq \Phi ^{\circ\mathbb{N}}(L_{A^1}L_{Nis} X)\simeq L_{A^1}L_{Nis} X$.

The first equivalence comes from that $\Phi^{\circ \mathbb{N}}$ preserves weak equivalences if $\Phi$ does, the second should come from applying $\Phi$ tranfinitely to $X\simeq \Phi (X)$. But if weak equivalences are already closed under transfinite composition. Why do they need to show that $\Phi^{\circ \mathbb{N}}X$ is $A^1$-fibrant?