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.