Every symplectic action is Hamiltonian

The displayed formula descends from the smooth translations $(x,y)\mapsto (x+t,y)$ of ${\mathbb{R}}^2$, and the identity and action laws hold by addition, so it is a smooth left action by [F2]. These translations preserve $dx$, $dy$ and hence $\omega =dx\wedge dy$, so the action is symplectic.

By [F2] the fundamental field for $\xi =1$ is ${\frac{d}{dt}\mid }_0(-t)\cdot p=-{\mathrm{\partial }}_x$. Thus the component equation would read $d{\mu }^1=-{\iota }_{-{\mathrm{\partial }}_x}\omega ={\iota }_{{\mathrm{\partial }}_x}\omega =dy$, so ${\mu }^1$ would be a primitive of $dy$.

But $dy$ has nonzero period: integrating it over the closed loop $\gamma (t)=[(0,t)]$, $0\le t\le 1$, gives $1$, while the integral of an exact one-form over a closed loop vanishes. Hence $dy$ is not exact, and no such function ${\mu }^1$ exists.

The translation action is therefore symplectic but not Hamiltonian, so the statement is false.