As an alternative to the other (perfectly fine) answer, we can also use the fact that an entire function can grow arbitrarily fast on the real line, and that functions generated by $S$ (including iterated integrals of those) can only grow as fast as finitely iterated exponentials.
In order to make the argument rigorous, let $e_1(x) = e^{x}$, and inductively $e_{k+1}(x) = e_k(e^x)$ be the iterated exponentials. Now define $ f(z)=a_0+\sum_{k=1}^\infty \left(\frac{z}{k}\right)^{n_k}$ where $a_0 = e_1(1)=e$, and $(n_k)$ is a strictly increasing sequence of natural numbers chosen such that $\left(\frac{k}{k-1}\right)^{n_{k-1}}\ge e_{k}(k)$ for all $k \ge 2$. Then $f$ is an entire function (since $|z/k|<1$ for $k > |z|$, the tail of the series is dominated by a geometric series for any fixed $z$), and $f(k) \ge e_k(k)$ for all $k\ge 1$ (since the $k$-th term in the series is $\ge e_k(k)$, and all other terms are positive.) It is easy to see by induction that each $e_n$ is increasing, and that $e_k(x) \ge e_m(x)$ for $k \ge m$ and all $x\ge 0$. This implies that $f(k) \ge e_k(k) \ge e_m(k)$ for all $k \ge m$, so $\limsup_{x\to\infty} \frac{f(x)}{e_k(x)} \ge 1$ for all $k$.
On the other hand, all functions $g \in S$ satisfy $\limsup_{x\to\infty} \frac{|g(x)|}{e_2(x)} = 0,$ and by induction any function $g$ that arises from functions in $S$ by $k$ operations (addition, multiplication, division, composition, integration) satisfies $\limsup_{x\to\infty} \frac{|g(x)|}{e_{k+2}(x)} = 0.$ This shows that $f$ is not in the class of entire functions generated by $S$.