Does Provable P equal Provable NP?
You need to define $L$ carefully. Currently, you seem to require only these properties:


"$L$ proves $[A \in \mathsf{P}]$" implies $A \in \mathsf{P}$
"$L$ proves $[A \in \mathsf{NP}]$" implies $A \in \mathsf{NP}$
"$L$ proves $[\mathrm{SAT} \in \mathsf{NP}]$"


Let $L$ be such that it ever proves only these two statements:


$[\mathrm{ONE} \in \mathsf{P}]$, where $\mathrm{ONE}$ is a dummy problem like "always accept", which is clearly in $\mathsf{P}$.
$[\mathrm{SAT} \in \mathsf{NP}]$


Then $\mathsf{ProvableP} = \{\mathrm{ONE}\}$ and $\mathsf{ProvableNP} = \{\mathrm{SAT}\}$, which are different even if $\mathsf{P}=\mathsf{NP}$. Note also that $\mathsf{ProvableP} \nsubseteq \mathsf{ProvableNP}$.