While your question as stated has been answered, I'd like to give a satisfying answer to a slight alteration of your question. To avoid the issue of commutivity, we can instead look at commutative hyperoperations $F_n(a,b)$. A few of them are:
\begin{align*} F_1(a,b)&=a+b=b+a\\ F_2(a,b)&=ab=ba\\ F_3(a,b)&=a^{\ln b}=b^{\ln a}\\ \end{align*}
Define $\ln^{(k)}(x):=\underbrace{\ln(\ln(\dots\ln(x)))}_{k\text{ times}}$. It is true that, in general:
$$\ln^{(k)}(F_n(a,b))=F_{n-k}(\ln^{(k)}(a),\ln^{(k)}(b))\tag{$*$}$$
for all integers $n\geq 1$ and $0\leq k < n$.
In other words, if we want to "go down" $k$ operations, then we need to use the $\ln$ function $k$ times.
Indeed, this aligns with the standard rule that $\ln(ab)=\ln a +\ln b$, or in our commutative hyperopration notation, $\ln(F_2(a,b)) = F_1(\ln a,\ln b)$.
So, for (my slight alteration of) your question, we want a function $f$ that satisfies
$$f(F_3(a,b))=f(a^{\ln b})=f(a)+f(b)=F_1(f(a),f(b)).$$
Here, $n=3$ and $k=2$, so we can set $f(x)=\ln^{(2)}(x)=\ln(\ln(x))$.
$$\ln(\ln(a^{\ln b}))=\ln(\ln(a)\cdot\ln(b))=\ln(\ln(a))+\ln(\ln(b))$$
Proof of $(*)$:
The commutative hyperoperations are defined recursively by
$$F_n(a, b) = \exp(F_{n-1}(\ln(a), \ln(b))).$$
Therefore,
$$F_n(a, b) = \exp^{(k)}(F_{n-k}(\ln^{(k)}(a), \ln^{(k)}(b))).$$
Hence,
$$\ln^{(k)}(F_n(a, b)) = F_{n-k}(\ln^{(k)}(a), \ln^{(k)}(b)).$$