If $f$ is a solution, it is injective, hence increasing or decreasing. If $f$ was surjective, $\mathrm{exp}$ would be too, so $f$ is not. If $f$ was decreasing, it would have a fixed point, and $\mathrm{exp}$ too, so $f$ is increasing. The set $f(\mathbb{R})$ is equal to $]a,b[$ for some $a, possibly infinite. If $b \neq + \infty$, $\exp (x) < b$ for all $x$, contradiction. So $b = + \infty$ and since $f$ is not surjective, $a \neq - \infty$. If $a \geq 0$, $\exp (x) \geq f(a)>0$ for all $x$, and that's impossible ($x=\log f(a)-1$ for example). So $a<0$. Since $\lim_{- \infty} f = a$, $\lim_{- \infty} \exp = f(a)$, and so $f(a)=0$.
Now the whole function $f$ can be reconstructed from $f|_{]- \infty, a]}$. Assume we only have a continuous, increasing and surjective function $f: ]-\infty,a] \rightarrow ]a,0]$. We can extend it to a continuous function on $\mathbb{R}$ such that $f \circ f = \exp$. For example, if $a < x \leq 0$, $x=f(y)$ for some unique $y \in ]-\infty,a]$, and so $f(x)=\exp y = \exp f^{-1}(x)$ ($f^{-1} : ]a,0] \rightarrow ]-\infty,a]$). It is easily checked that this defines a continuous, increasing and surjective function $f : ]-\infty,0] \rightarrow ]a,\exp a]$ (only the continuity at $a$ is not completely obvious, but easy). We can go on to extend $f$ on $]0,\exp (a)]$, $]\exp (a),1]$, etc.
So there are infinitely many continuous solutions, and a recipe that gives them all: pick $a<0$, and "pick" a continuous, increasing and surjective function $f : ]-\infty,a] \rightarrow ]a,0]$. There is a unique continuous extension of $f$ such that $f \circ f = \exp$.
If you want $f$ to be $C^k$, $C^{\infty}$, analytic; you need to take $f$ that is so on $]-\infty,a]$, and check that it is so around $a$ when you do the first extension (the behavior of $f$ on the right of $a$ is controlled by its behavior at $-\infty$, because $f(a+\epsilon)=\exp f^{-1} (a+\epsilon)$). Have fun!