= Solution
In one canonical degree of freedom it is enough to check $\{Q,P\}=1$ for a locally invertible transformation. Here
$$
\{Q,P\}=ff'(\cosh^2q-\sinh^2q)=ff',
$$
so $ff'=1$ and $f(p)^2=2p+C$. To obtain exactly the requested <Hamiltonian>, choose $C=0$ and the positive branch for $p>0$:
$$
\boxed{f(p)=\sqrt{2p},\qquad
Q=\sqrt{2p}\sinh q,\quad P=\sqrt{2p}\cosh q.}
$$
The identities $\cosh^2q-\sinh^2q=1$ and $\cosh^2q+\sinh^2q=\cosh2q$ give
$$
\boxed{\frac{Q^2+P^2}{2}=p\cosh2q=H.}
$$
This is a local canonical chart, with inverse
$$
p=\frac{P^2-Q^2}{2},\qquad q=\operatorname{artanh}(Q/P),\qquad P>|Q|.
$$
It is not a global identification of the real $(q,p>0)$ half-plane with the entire oscillator plane. The negative branch for $f$ gives the analogous lower wedge. This domain matters if an oscillator trajectory is continued beyond the boundary of the chart.
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