Does the inclusion $A \subset B$ for closed densely defined operators $A$ and $B$ imply $A=B$?

If $A : \mathcal{D}(A)\subset\mathcal{H}\rightarrow\mathcal{H}$ is a densely-defined, closed, symmetric linear operator on a complex Hilbert space $\mathcal{H}$, then the adjoint $A^*$ is a densely-defined closed linear operator, and one has the graph inclusion $\mathcal{G}(A)\subseteq\mathcal{G}(A^*)$, and the following orthogonal decomposition in $\mathcal{H}\times\mathcal{H}$:

$$ \mathcal{G}(A^*)=\mathcal{G}(A)\oplus\mathcal{D}_{-}\oplus\mathcal{D}_{+}, $$

where $\mathcal{D}_{\pm}$ are the restrictions of $A^*$ to $\mathcal{N}(A^*\pm iI)$, respectively. So, if $A$ is symmetric, but not selfadjoint, then $A$ has a proper closed extension $A^*$.

For example, define the operator $Lf=-if'$ on $\mathcal{D}(L)\subset L^2[0,1]$ consisting of functions $f \in L^2[0,1]$ that are equal a.e. to absolutely continuous functions $f$ with $f'\in L^2$ and $f(0)=f(1)=0$. $L : \mathcal{D}(L)\subset L^2\rightarrow L^2$ is a closed, densely-defined, symmetric operator that is not selfadjoint. The domain of $L^*$ is the same as that of $L$, except without the endpoint restrictions. Both $L$ and $L^*$ are closed and densely-defined, and $L^*$ is a proper extension of $L$.