No. Take $f(x) = 2 \pi i x$.

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There is also a more subtle way I can cheaply answer the question: It is easy to give functions $f(x)$ such that the Newton series of $f$ converges to $f$, but the Newton series of $e^f$ diverges. Take $f(x)=x^2$ or, more subtly, $f(x) = \cos (2 \pi x/8)$. I assume that the right formulation of the question is "If $f(x)$ is real, the Newton series of $f$ converges to $f$, and the Newton series of $e^f$ converges, does the Newton series of $e^f$ converge to $e^f$?