Is the breaking of PT symmetry in a quantum mechanical system indicating the transition from a non-dissipative to a dissipative system: no, PT symmetric systems are in between closed (Hermitian) and open (general, non-Hermitian) systems in the following sense.
A quantum mechanical system with a Hermitian Hamiltonian is closed and probability is conserved (and in its classical analog (if it exists), the total energy in the system is conserved). A QM system with non-Hermitian Hamiltonian generally is open, it looses probability and indeed exhibits dissipation in this sense.
A quantum mechanical system with a non-Hermitian, PT-symmetric Hamiltonian may exhibit PT symmetric eigenfunctions and real eigenvalues (unbroken PT symmetry) or complex eigenvalues and non-PT symmetric eigenfunctions.
As a simple example, we consider two subsystems one of which loses probability (and the classical counterpart would lose energy, so there is dissipation) and the other gains exactly the same amount. Such coupled QM system is PT symmetric and exhibits real eigenvalues for sufficiently large coupling strength such that the eigenfunctions are also PT symmetric (unbroken PT symmetry). If the coupling becomes too weak, the eigenvalues cease to be real but become complex, indicating a state of broken PT symmetry, i.e. eigenfunctions are not any longer PT symmetric.
The breaking of PT-symmetry, rather than a transition from an open (dissipative) to a closed (non-dissipative) system, indicates a transition from a balance between loss and gain to a state without such balance.