There is no entire (holomorphic everywhere) function $f(z)$ with $f(f(z)) = e^z$. To see this, one can use Picard's little theorem, and do case analysis. For example, $f(z)$ cannot take all values, for then so does $f(f(z))$, while $e^z$ omits zero. The case that $f(z)$ has an omitted value can also be excluded, this is not difficult.

Alternatively, use the theorem of Polya that if $f(z)$ and $g(z)$ are entire functions, then 
$f(g(z))$ is of infinite order unless (i) $f(z)$ is of finite order and $g(z)$ is a polynomial, or (ii) $f(z)$ has order zero and $g(z)$ is of finite order. The exponential function has order $1$.