The annihilators in dual operator algebras.

It is well known that in every von Neumann algebra $\mathcal{M}$, the left annihilator of a given subset $S\subseteq \mathcal{M}$ is in the form $\mathcal{M}p$ for some projection $p$ in $\mathcal{M}$.

By a dual operator algebra $\mathcal{A}$, we mean a WOT-closed subalgebra of operators on some Hilbert space $H$.

Q. Does there exist any similar characterization of the left annihilators in dual operator algebras?

The result you quote holds more generally:

If $\mathscr{L}$ is a left $\sigma$-weakly closed ideal in a von Neumann algebra $\mathscr{M}$, then $\mathscr{L}= \mathscr{M}e$ for a unique projection $e \in \mathscr{L}$.

Note that every left annihilator is a $\sigma$-weakly closed left ideal.

We still have the following:

If $\mathscr{L}$ is a left $\sigma$-weakly closed ideal in a 'dual operator algebra' $\mathscr{M}$, then $\mathscr{L}= \mathscr{M}e$ for a unique projection $e \in \mathscr{L}$.

Proof: The proof of the classical result from Takesaki's first volume (chapter II, proposition 3.12) carries over without modification. For convenience, I copy it here in my own wording.

Consider $\mathscr{N}:=\mathscr{L} \cap \mathscr{L}^*$. Then $\mathscr{N}$ is a $\sigma$-weakly closed $*$-subalgebra of $\mathscr{B}(H)$. Hence, $\mathscr{N}$ is $\sigma$-strong$^*$-closed as well and by theorem 3.9 in chapter II of Takesaki's first volume, there exists a greatest projection $e \in \mathscr{N}$ which is a unit for $\mathscr{N}$. We show that $\mathscr{M}e = \mathscr{L}$ and we will be done.

Clearly, $\mathscr{M}e\subseteq \mathscr{L}$ since $e \in \mathscr{L}$ and $\mathscr{L}$ is a left ideal. Conversely, assume $x \in \mathscr{L}$. Then $$x^*x = (x^*x)^* \in \mathscr{L} \cap \mathscr{L}^* = \mathscr{N}$$ so also $|x|:= (x^*x)^{1/2} \in \mathscr{N}$ since $\mathscr{N}$ is a $C^*$-algebra and $C^*$-algebras are closed under taking square roots of positive elements. Since $e$ is a unit for $\mathscr{N}$, we have $|x| = |x|e$ and if $x=u|x|$ is the polar decomposition of $x$ (so $u$ is a partial isometry satisfying $\ker(u) = \ker(x))$, we get $$xe = u|x|e = u|x| = x$$ so $x= xe \in \mathscr{M}e$, which shows $\mathscr{L}\subseteq \mathscr{M}e$.

Uniqueness of the projection is trivial and this ends the proof. $\quad \square$