Sums of entire surjective functions

One expects there to be no such $a_n$ in general, because the "typical" entire functions is surjective (those that aren't are of the special form $z \mapsto c + \exp g(z)$). An explicit example is $f_n(z) = \cos z/n$: any convergent linear combination $f = \sum_n a_n f_n$ is of order $1$, so if $f$ is not surjective then $g$ is a polynomial of degree at most $1$; but $f$ is even, so must be constant, from which it soon follows that $a_n=0$ for every $n$.

The answer is no. If something does not hold for polynomials, don't expect that it will hold for entire functions:-)

For example, all non-constant functions of order less than $1/2$ are surjective. This follows from an old theorem of Wiman that for such function $f$ there exists a sequence $r_k\to\infty$ such that $\min_{|z|=r_k}|f(z)|\to\infty$ as $k\to \infty.$ And of course linear combinations of functions of order less than $1/2$ are of order less than $1/2$.

Edit. To construct a counterexample with infinite sums, one can use lacunary series. Let $\Lambda$ be a sequence of integers $n_k$ which grows sufficiently fast, for example, such that $n_k/k\to\infty$, and consider the class of entire functions of the form $$f(z)=\sum_{n\in\Lambda}c_nz^n.$$ It is known that all such functions are surjective. And of course any linear combination of such functions, finite or infinite, belongs to the class.

Reference: L. Sons, An analogue of a theorem of W.H.J. Fuchs on gap series, Proc. LMS, 1970, 21 525-539.