I enjoyed Christian's answer, and if you want the shortest proof starting from the series definitions, then that is the way to go. But, like you, I was curious about this question because I like to manipulate power series, and because this problem ended up being harder than I thought it would be.
Here is a solution that meets your requirements. The only thing I have omitted is proving that exchanging the order of summation is valid when I do it below, which is probably not the part of the proof you are interested in.
A few lemmas are needed, which I will list first. The first two are well-known combinatorial identities, and I will omit the proofs. I'll follow your approach, but I'll delay the expansion of $e^x$ a little bit longer than you did, because otherwise you get sums over an arbitrarily large number of variables, which end up being difficult to handle.
Lemma 1: For $m,n \in \mathbb{Z}$ with $0 \le m < n$,
$$ \sum_{k=0}^n (-1)^k \binom{n}{k} k^m = 0 $$
The easiest way to prove this is with calculus, but it can be proven by induction on $m$ from the result that for $n \ge 1$,
$$ \sum_{k=0}^n (-1)^k \binom{n}{k} = 0 $$
which itself can be proven by induction on $n$. So, no calculus required.
Lemma 2:
$$ \sum_{n=k}^m \binom{n}{k} = \binom{m+1}{k+1} $$
This can also be proven by induction.
Lemma 3: $ e^x e^y = e^{x+y} $
Proof: This one is a lot easier than the main result. Expand the right hand side using the series definition and the binomial theorem to get
$$ e^{x+y} = \sum_{m=0}^\infty \frac{(x + y)^m}{m!} = \sum_{m=0}^\infty \sum_{i=0}^m \frac{1}{m!} \binom{m}{i} x^i y^{m-i} = \sum_{i=0}^\infty \sum_{m=i}^\infty \frac{x^i y^{m-i}}{i! (m-i)!} $$
Let $j=m-i$ and the sums become independent:
$$ \sum_{i=0}^\infty \sum_{m=i}^\infty \frac{x^i y^{m-i}}{i! (m-i)!} = \sum_{i=0}^\infty \sum_{j=0}^\infty \frac{x^i y^j}{i!j!} = \left( \sum_{i=0}^\infty \frac{x^i}{i!} \right) \left( \sum_{j=0}^\infty \frac{x^j}{j!} \right) = e^x e^y $$
Lemma 4: $ (e^x)^y = e^{xy} $ when $y$ is a nonnegative integer.
Proof: This is obvious using Lemma 3.
Main result: $ \log(e^x) = x $
Proof: Apply the power series for the logarithm, then a binomial expansion, and then the power series for the exponential.
$$ \begin{align}
\log(e^x) &= \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} ( e^x - 1 )^n \\
&= \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} \sum_{k=0}^n \binom{n}{k} (-1)^{n-k} e^{kx} \\
&= \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} \sum_{k=0}^n \binom{n}{k} (-1)^{n-k} \sum_{m=0}^\infty \frac{k^m x^m}{m!} \\
&= -\sum_{m=0}^\infty \frac{x^m}{m!} \sum_{n=1}^\infty \frac{1}{n} \sum_{k=0}^n \binom{n}{k} (-1)^k k^m \\
&= -\sum_{m=0}^\infty \frac{x^m}{m!} a_m \tag{1} \label{1}
\end{align} $$
where $a_m$ has been defined as
$$ a_m = \sum_{n=1}^\infty \frac{1}{n} \sum_{k=0}^n \binom{n}{k} (-1)^k k^m $$
Split the sum over $n$ into two parts, from $1$ to $m$ and from $m+1$ to $\infty$.
$$ a_m = \sum_{n=1}^m \frac{1}{n} \sum_{k=0}^n \binom{n}{k} (-1)^k k^m + \sum_{n=m+1}^\infty \frac{1}{n} \sum_{k=0}^n \binom{n}{k} (-1)^k k^m $$
On the right, the summation over $k$ is zero by Lemma 1 because $n > m$. This gives us
$$ a_m = \sum_{n=1}^m \frac{1}{n} \sum_{k=0}^n \binom{n}{k} (-1)^k k^m \label{2} \tag{2} $$
In this expression, when $k \ge 1$, we can apply the identity
$$ \frac{1}{n} \binom{n}{k} = \frac{1}{k} \binom{n-1}{k-1} \label{3} \tag{3} $$
Fortunately, the $k=0$ terms in \eqref{2} are all zero because of the $k^m$ term (since we are only considering $a_m$ where $m \ge 1$). Dropping the $k=0$ terms, moving the summation over $n$ to the right, and applying \eqref{3} in \eqref{2}, we get
$$ \begin{align}
a_m &= \sum_{k=1}^m (-1)^k k^m \sum_{n=k}^m \frac{1}{k} \binom{n-1}{k-1} \\
&= \sum_{k=1}^m (-1)^k k^{m-1} \sum_{n=k}^m \binom{n-1}{k-1}
\end{align} $$
Applying Lemma 2 to the summation over $n$ gives
$$ a_m = \sum_{k=1}^m (-1)^k k^{m-1} \binom{m}{k} $$
This can be simplified by again applying Lemma 1. However, note that Lemma 1 requires the $k=0$ terms, which we dropped earlier so that we could apply \eqref{3}. Now the $k=0$ term in the sum must be replaced by including $k=0$ in the sum and subtracting it outside the sum.
$$ a_m = \left( \sum_{k=0}^m (-1)^k k^{m-1} \binom{m}{k} \right) - 0^{m-1} $$
Obviously this term on the right is zero when $m>1$, but the $m=1$ case results in $0^0$. Here the appropriate convention is that $0^0=1$. (For example, when we expanded the exponential as $e^{kx} = \sum_m k^m x^m/m!$, the $m=0$ term is interpreted as $1$ even when $k=0$.) Since the $0^{m-1}$ term yields 0 for $m \ne 1$ and 1 for $m=1$, we can write this using the Kronecker delta as $\delta_{m,1}$.
$$ a_m = \left( \sum_{k=0}^m \binom{m}{k} (-1)^k k^{m-1} \right) - \delta_{m,1} $$
Now Lemma 1 can be applied again. (Note that the conditions of the lemma are satisfied because $m-1 < m$.) This gives $ a_m = -\delta_{m,1} $. Inserting this into $\eqref{1}$ gives the final result,
$$ \log(e^x) = x $$
