Let the real Hilbert space $H^1(\Omega)$ endowed with its usual inner product, denoted by $\langle ., . \rangle$
and let $A : H^1(\Omega) \rightarrow H^1(\Omega)$ be a dissipative operator ($\langle A y,y\rangle\leq 0, \; \forall y\in H^1(\Omega)$)
and $B : H^1(\Omega) \rightarrow H^1(\Omega)$ be a positive operator ($\langle B y,y\rangle\geq 0, \; \forall y\in H^1(\Omega)$).
Can we say that $BA$ is dissipative in other words ($\langle BA y,y\rangle\leq 0, \; \forall y\in H^1(\Omega)$)?
No. For a finite dimensional example let $H = \mathbb{R}^2$ and take
$$ A = - B = \begin{pmatrix} 0 & 1 \\ -1 & 0\end{pmatrix} \tag{exa}$$
For every $y$ you have $\langle Ay,y\rangle = 0 = \langle By,y\rangle$ so both are dissipative. But $AB = BA = I$.
For higher dimensions just take a two dimensional subspace, and let $A, B$ act trivially on the rest.
Even if you require strict inequality the statement is not true. Replace the $0$ in (exa) above by $-\epsilon$, and you have $A$ is strictly dissipative, $B$ is strictly positive, and $BA = I + O(\epsilon)$ is strictly positive.