The non-existence of a holomorphic compositional square root of the exponential can be proved by an elementary argument, which does not used Picard's theorem.

The range of $h$ contains that of $\exp$, so is either ${\mathbb C}$ or ${\mathbb C}^\times$. If $h$ was onto, then $h\circ h$ would be onto, which is false. Therefore $h({\mathbb C})={\mathbb C}^\times$.

The function $h'/h$ is thus entire. It admits a (unique) primitive $g$ with a given initial condition $g(0)$ such that $h(0)=\exp g(0)$. Then $(he^{-g})'\equiv0$ gives $h=\exp\circ g$.

We have $\exp\circ g\circ \exp\circ g=\exp.$ Hence, for every $z\in {\mathbb C}$, there exists an integer $k(z)$ such that $g\circ\exp\circ g(z)=z+2ik(z)\pi$. The function $z\mapsto 2ik(z)\pi=g\circ\exp\circ g(z)-z$ is continuous from ${\mathbb C}$ (connected) to $2i\pi{\mathbb Z}$ (discrete): it is constant.

Let $T$ denote the translation $z\mapsto z+2ik\pi$, which is a bijection. From $g\circ(\cdots)=T$, we see that $g$ is onto. From $(\cdots)\circ g=T$, we see that $g$ is one-to-one. Hence $g$ is bijective. 

Finally, $\exp=g^{-1}\circ T\circ g^{-1}$ is bijective, an obviously false statement.

**Edit**. Actually, the proof works almost the same for a continuous compositional square root. We only need to prove the existence of a continuous $g:{\mathbb C}\rightarrow{\mathbb C}$ such that $h=\exp\circ g$. This is guaranted by the fact that $\exp:{\mathbb C}\rightarrow{\mathbb C}^\times$ is a covering with ${\mathbb C}$ simply connected (universal cover).