Assumption: A PSPACE-hard problem can be solved in polynomial time
Your argument is perfectly fine. However, you actually do not even have to argue with polynomial reduction.
Since $\textsf{P} \subseteq \textsf{NP}, \textsf{coNP} \subseteq \textsf{PSPACE}$, under the assummption that there is a $\textsf{PSPACE}$-hard problem solvable in polynomial time (i.e. it is in $\mathsf{P}$), you also have $\textsf{PSPACE} \subseteq \mathsf{P}$. Now $\textsf{NP}, \textsf{coNP}, \textsf{PSPACE}$ are sandwiched in $\mathsf{P}$ and thus, equality holds for the $\subseteq$ relations above.