The theorem I am referring to is from "The Concrete Tetrahedron" by Kauers and Paule, which states:
Theorem 2.8 (Transfer Principle) Let $a(z) = \sum_{n=0}^{\infty} a_nz^n$ and $b(z) = \sum_{n=0}^{\infty} b_nz^n$ be real or complex functions analytic in a non-empty open neighbourhood of zero. If $a(z) = b(z)$ for all $z \in U$, then $a_n = b_n$ for all $n \in \mathbb{N}$.
An example given in the book to illustrate this principle is:
"The transfer principle can be used for obtaining simple proofs of identities. For instance, to demonstrate that $$ \exp(\log(1+x)) = 1 + x $$ as formal power series, it is sufficient to observe that this relation is true for the corresponding analytic functions. Proving this without the transfer principle would necessitate elaborate calculations."
My interpretation is that the Transfer Principle implies that the Taylor series of the analytic function $\exp(\log(1+x))$ is simply $1 + x$.
However, when we regard $\exp(\log(1+x))$ as a formal power series (generating functions), we use the definitions: $$ \exp(x) = \sum_{n=0}^\infty \frac{x^n}{n!} $$ and $$ \log(1+x) = \sum_{n=1}^\infty \frac{(-1)^{n+1}x^n}{n}. $$ Then, by the definition of the composition of power series, we get: $$ \exp(\log(1+x)) = \sum_{n=0}^\infty \frac{\left(\sum_{n=1}^\infty \frac{(-1)^{n+1}x^n}{n}\right)^n}{n!}. $$ My confusion arises here: How can we assert that the above power series, $$ \sum_{n=0}^\infty \frac{\left(\sum_{n=1}^\infty \frac{(-1)^{n+1}x^n}{n}\right)^n}{n!}, $$ is equivalent to the Taylor series of $\exp(\log(1+x))$ and therefore equals $1+x$?
I am unable to see how this conclusion is derived directly from the Transfer Principle.
Update: I don't have a problem with the theorem itself. But I am not sure it implies the identity given as an example. More specifically, if we consider $\exp(x)$ and $\log(1+x)$ as short-hands to write the corresponding power series, how do we know that $$ \exp(\log(1+x)) $$ as a composition of two power series, has the same coefficients as $$ \exp(\log(1+x)) $$ when considered as a composition of two analytic functions.
From what you have written here, you are definitely right: At some point one has to argue (or at least mention) that the two types of composition are the same. I took a quick look at the referenced book and indeed I could not find it there.
However, you can deduce this quite directly from the definition of convergence and composition of formal power series as defined on page 24 and 25. I added the relevant parts below.
From the construction of the composition it is clear that the coefficients of the composed formal power series are the same as if one would compose the formal power series as analytic functions.