The answer to your original question is NO. Here is an counterexample:

For simplicity, assume everything is real-valued and the Hilbert space is over $\mathbb{R}$, let $$ \varphi \in L^2(\mathbb{R}) \setminus L^\infty(\mathbb{R}) \text{ and } \varphi\chi_{[0,1]}\in L^\infty([0,1]);$$
$$\psi \in L^2([0,1]) \cap L^\infty([0,1]).$$

Consider the rank one operator $A = \varphi \otimes \psi$, i.e.,
$$
A(\xi) = \varphi \langle \xi, \psi\rangle_{L^2(\mathbb{R})}.
$$
Then $A$ verifies of course your assumption. Indeed, we have
$$
Af = \varphi \otimes (\psi\cdot f) \text{ and } \mathrm{tr} (Af) = \int\varphi(x) \psi(x) f(x) dx \le \|\varphi\psi \|_{L^2([0,1]} \| f\|_{L^2([0,1])} \le C\| f\|_H.
$$
But now, for $fA$ to be in $B(H)$, it is necessary that for any $\xi \in H$, we have
$$
f \in H \Longrightarrow f \cdot A(\xi) \in H,
$$
this means that $A(\xi) \in L^\infty(\mathbb{R}) \cap L^2(\mathbb{R})$. But by our choice of $\varphi$, we know that
$$
A(\psi) = \| \psi\|_H^2\cdot \varphi \notin L^\infty(\mathbb{R}) \cap L^2(\mathbb{R}).
$$
This means that $fA$ is not even in $B(H)$.