tr(ab) = tr(ba)?

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It is well known that given two Hilbert-Schmidt operators $a$ and $b$ on a Hilbert space $H$, their product is trace class and $tr(ab)=tr(ba)$. A similar result holds for $a$ bounded and $b$ trace class.

The following attractive statement, however, is false:

Non-theorem:
Let $a$ and $b$ be bounded operators on $H$. If $ab$ is trace class , then $ba$ is trace class and $tr(ab)=tr(ba)$.

The counterexample is $a=\pmatrix{0&0&0\\0&0&1\\0&0&0}\otimes 1_{\ell^2(\mathbb N)}$, $b=\pmatrix{0&1&0\\0&0&0\\0&0&0}\otimes 1_{\ell^2(\mathbb N)}$.

I’m guessing that the following is also false, but I can’t find a counterexample:

Non-theorem?:
Let $a$ and $b$ be two bounded operators on $H$. If $ab$ and $ba$ are trace class, then $tr(ab)=tr(ba)$.