Suppose that $A$ is a Banach algebra satisfying the condition $x\in A, x\cdot A=\{0\}\Rightarrow x=0.$ Given a nonempty subset $C\subset A$ we define the left annihilator $\ell(C):=\{x\in A : x\cdot C=\{0\}\}$.

Let $B$ be a closed subalgebra $B$ of $A$. Is it true that $\ell(B)=\{0\}\Rightarrow B=A$?

ADDENDUM: I am interested in the cases $A=L_1(G)^{**}$ and $A=L_1(G)^{**}/L_1(G)$, where $G$ is a compact abelian group, $L_1(G)$ is the convolution algebra, and $L_1(G)^{**}$ is endowed with the first Arens product. In this case $L_1(G)$ is an ideal in $L_1(G)^{**}$, and both examples satisfy the required condition because $L_1(G)^{**}$ admits a right identity $E$ ($x\cdot E= x$ for each $x\in A=L_1(G)^{**}$).