Does there exist a function which converts exponentiation into addition?

If $f(a)+f(b)=f(a^b)$ for all $a,b\in\mathbb R^+$, then it follows (by interchanging $a$ and $b$) that $f(b)+f(a)=f(b^a)$ for all $a,b\in\mathbb R^+$. Hence, $f(a^b)=f(b^a)$ for all $a,b\in\mathbb R^+$. Setting $b=1$, we see that $f(a)=f(1)$ for all $a\in\mathbb R^+$. Now $f(1)+f(1)=f(1^1)$, so $f(1)=0$. Hence, the only function $\mathbb R^+\to\mathbb R$ with the desired property is the zero function.

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, $k=2$, 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,

$$\ln^{(k)}(F_n(a, b)) = F_{n-k}(\ln^{(k)}(a), \ln^{(k)}(b)).$$