 Solution

For the standard symplectic form, $D_f={\sum }_a(f_{p_a}{\partial }_{q^a}-f_{q^a}{\partial }_{p_a})$; a transformation is canonical when it preserves the symplectic form, equivalently Poisson brackets. The Jacobi identity gives $[D_f,D_g]=D_{\{f,g\}}$ with the stated convention. For $Z^A=e^{-s{D}_f}{z}^A$, differentiate $\{Z^A,Z^B\}$ in $s$; Jacobi makes the derivative the Lie transport of the bracket, preserving the canonical constant brackets at $s=0$. Hence every map in the one-parameter flow is canonical.