Take $S$ to be the set containing all polynomials and $e^z$ over $\mathbb C$. If we add, subtract, multiply, divide (if the denominator is non-zero everywhere) and compose functions in our set $S$ we get analytic functions.
My questions is: do all analytic functions on the whole complex plane arise this way?
$\int_0^z e^{w^2} dw$ is an entire function which is not elementary, so no.
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 $S$.
