Show that there does not exist a holomorphic function $h(z)$ such that $\exp(h(z)) = z$ on the punctured plane without using complex integration?

Let's answer the question in a more general context: There exists no continuous function $h:\mathbb{C}\setminus\{0\}\rightarrow\mathbb{C}$ such that $\exp \circ h = id\vert_{\mathbb{C}\setminus\{0\}}$. (Simply connected means that I can change every loop continuously into the null loop, i.e. a single point.)

For the proof, we need some concepts about loops and fundamental groups. Explain everything in detail would make this post into a script, so I will just give the ideas. If you draw a loop starting from (end thus ending in) some $z_0\in \mathbb{C}$, then you can continuously deform it into a single point. However, if you consider $\mathbb{C}\setminus\{0\}$, this is not the case anymore. Indeed, if a loop "goes around $0$, then it is impossible to continuously deform it into a point. By saying that two loops are equivalent iff we can continuously deform one into the other, we get an equivalence relation. In the example of $\mathbb{C}\setminus\{0\}$, two loops are equivalent iff they "go around $0$" the exact same number of times (in the same direction, i.e. clockwise or counter-clockwise). More generally, the equivalence classes of loops form a group with respect to the concatenation of loops. If we look at some (connected) open set $U\subseteq\mathbb{C}$ and some $z_0\in U$, then we write $\pi(U;z_0)$ for this group and we call it the fundamental group. This group describes "the holes" in $U$ in some sense. For example, $\pi(\mathbb{C}\setminus\{0\}, z_0) = \mathbb{Z}$, because we count, how many times a loop "goes around $0$" (positive for the counter-clockwise direction). [It is possible to define this notion in a much more general context!] As for any good structural description, we can translate it when we have a suitable transformation between to set $U,V\subseteq\mathbb{C}$. Indeed, if we have a continuous function $f:U\rightarrow V$, we may define $f(\alpha)$ for any loop $\alpha$ in $U$ as the image of the loop under $f$. Since $f$ is continuous, if $\alpha$ and $\beta$ are equivalent, then so are $f(\alpha)$ and $f(\beta)$. That means that we may define the function $\pi_f: \pi(U,z_0)\rightarrow\pi(V, f(z_0))$ through $\pi_f([\alpha]) := [f(\alpha)]$, where $[x]$ denotes the equivalence class of $x$. This is all we need, I hope it was not too confusing.

Suppose now there were such a function $h$. Since it is continuous, we may consider $\pi_h : \pi(\mathbb{C}\setminus\{0\}, z_0) \rightarrow \pi(\mathbb{C}, h(z_0))$. The same goes for $\pi_{\exp} : \pi(\mathbb{C}, h(z_0))\rightarrow \pi(\mathbb{C}\setminus\{0\}, z_0)$ and $\pi_{\exp\circ h} = \pi_{id_{\mathbb{C}\setminus \{0\}}} = id_{\mathbb{Z}} : \pi(\mathbb{C}\setminus\{0\}, z_0)\rightarrow \pi(\mathbb{C}\setminus\{0\}, z_0)$. It is easy to see that $$\pi_{\exp\circ h} = \pi_{\exp}\circ \pi_h$$. But since $\pi(\mathbb{C}, h(z_0)) = \{ 0\}$ (all loops can be continuously deformed to one point), we have $\pi_{\exp} = 0$ and thus $$ id_{\mathbb{Z}} = \pi_{\exp\circ h} = 0, $$ which is a contradiction.

You see that the proof is rather short if the basics are known. And it is worth while to take a closer look at fundamental groups. They are amazing objects and I wish I would have learned more about them before deviating to other domains... They initially came from Poincaré if I am not mistaken and they "characterise" pointed topological spaces. This website has some very nice and easy to follow explanations (and visualisations). Unfortunately, it is completely in French, but that is a beautiful language to learn anyway! (I am not French.)

Suppose such an $h(z)$ exists. Without loss of generality, we may select $h(1) = 0$ (the argument is similar for any other permissible choice of $h(1)$). Then for any $z, w \in \Omega := \Bbb{C} \setminus \{0 \}$, we must have $$e^{h(z)} e^{h(w)} = zw = e^{h(zw)},$$ which immediately implies the identity $h(zw) = h(z) + h(w) + 2\pi i k(z, w)$ for some continuous function $k: \Omega \times \Omega \rightarrow \Bbb{Z}$.

Since $\Omega \times \Omega$ is connected, $k(z, w)$ must be a constant $k$, and plugging in $z = w = 1$, we find $h(1) = 2h(1) + 2\pi i k$, so that in fact $k = -h(1)/2\pi i = 0$ and $h(zw) = h(z) + h(w)$ for all $z, w \in \Omega$.

By induction we get $$h(z_1 ... z_m) = h(z_1) + ... + h(z_m), \text{ for all } z_1, ..., z_k \in \Omega.$$

If we let $z_1 = z_2 = z_3 = -1$, then the above identity shows $h(-1) = 3h(-1)$, which is only possible if $h(-1) = 0$. But $e^0 = 1 \neq -1$, contradiction.

Edited to add: This type of approach often works well for showing that other inverse functions or multivalued functions (e.g. the square root $f(z) = \sqrt{z}$) cannot be extended holomorphically to $\Bbb{C}$ or $\Bbb{C} \setminus \{ 0 \}$. One can play with a functional equation or identity such a function must satisfy, and then derive a contradiction from the function needing to take on two different values at the same point (without having to use path or contour integration).

To show that the square root $f(z) = \sqrt{z}$ cannot be continuously extended to all of $\Bbb{C}$, for instance, one establishes that $f(zw) = \alpha(z, w) f(z)f(w)$, where $\alpha(z, w) \in \{ \pm 1 \}$; uses connectedness of $\Bbb{C}^2$ to establish that $\alpha(z, w)$ is constant for all $(z, w) \in \Bbb{C}^2$ (either $+1$ or $-1$); and then derives a contradiction from the fact that $\alpha(-1, -1)$ must have the opposite sign as $\alpha(+1, +1)$.(E.g. $f(1) = 1$, then $\alpha(1, 1) = 1$ but $\alpha(-1, -1) = -1$.)

There cannot be a continuous function $h$ on the unit circle $|z|=1$ such that $e^{h(z)}=z$. If there were, then a connectedness argument could be applied to prove that $h(e^{i\theta})=i\theta+2n\pi i$ for all $0 \le \theta < 2\pi$ and for some fixed integer $n$, which would contradict the assumed continuity of $h$ at $z=1$.