There is a much more general result of Vallée-Poussin from which a negative answer to your question follows.
Let $(X,\mu)$ be a measure space. We say that a family of function $\mathcal{F}\subset L^1(X)$ is equi-integrable if for every $\varepsilon>0$ there is $\delta>0$ such that $$ \sup_{f\in\mathcal{F}} \int_E |f|\, d\mu<\varepsilon \quad \text{whenever } \mu(E)<\delta. $$
Theorem (de la Vallee Poussin). Let $(X,\mu)$ be a measure space with $\mu(X)<\infty$ and let $\mathcal{F}\subset L^1(X)$. Then $\mathcal{F}$ is equi-integrable if and only if there is a Young function $\Phi$, $\lim_{t\to\infty}\Phi(t)/t=\infty$ such that $$ \sup_{f\in\mathcal{F}}\int_X\Phi(|f|)\, d\mu\leq 1. $$
Clearly a family consisting of a single function is equi-integrable (by absolute continuity of the integral) so for each $f\in L^1$ there is $\Phi$ with $\Phi(f)\in L^1$.
Also we can adjust the function $\Phi$ around zero arbitrarily without changing integrability of the function $\Phi(f)$ so you can have the condition $\Phi(t)/t\to 0$ as $t\to 0^+$.
You can find the proof of the de la Vallee Poussin theorem in
C. Dellacherie, P.-A. Meyer, Probabilities and potential. C. Potential theory for discrete and continuous semigroups. Translated from the French by J. Norris. North-Holland Mathematics Studies, 151. North-Holland Publishing Co., Amsterdam, 1988.
M.M. Rao, Z.D, Ren, Theory of Orlicz spaces. Monographs and Textbooks in Pure and Applied Mathematics, 146. Marcel Dekker, Inc., New York, 1991.
