Von Neumann algebra generated by a non-self-adjoint operator

Let $$A$$ be a bounded linear operator on a separable Hilbert space $${\cal H}$$, and suppose that $$A$$ is distinct from its adjoint $$A^*$$.

Question: Can the double commutant of $$A$$ be distinct from the double commutant of $$\{A,A^*\}$$? If so, is there a simple example?

This question was inspired by the following statement from section 3.3 in Vaughan Jones (2009), Von Neumann Algebras (https://math.berkeley.edu/~vfr/VonNeumann2009.pdf):

If $$S \subseteq {\cal B(H)}$$, we call $$(S \cup S^*)''$$ the von Neumann algebra generated by $$S$$.

I don't know if this was meant to be the most efficient definition, and that's exactly what prompted my question. Maybe the Hilbert space wasn't assumed to be separable in that context, but I am interested in the separable case (if it matters).

Here is a simple example. Let $$H=\mathbb C^2$$ and $$A=\begin{bmatrix}0&1\\0&0\end{bmatrix}$$. Then $$\{A,A^*\}''=M_2(\mathbb C)$$, while $$\{A\}''=\left\{\begin{bmatrix}a&b\\0&a\end{bmatrix}:\ a,b\in\mathbb C\right\}.$$
Turner studied weighted (two-sided) shifts on the Hilbert space $$\ell_2(\mathbb Z)$$. He proved the following theorem (see Theorem 3.4 here)
Theorem. A two-sided shift on $$\ell_2(\mathbb Z)$$ is not invertible if and only if it is in the class (dc).
It may actually happen that an operator $$T$$ is in the class (dc) but $$T\oplus 0$$ is not (see Section 4 of the linked paper).