There is no analytic local solution at $0$ to $f'=e^{f^{-1}}$, $f(0)=0$, that is, the formal power series solution is diverging. Together with the solution given in comments by fedja, this means the actual scenario is **2.** For convenience of notation, I shall consider the equivalent equation
$$\begin{cases} g' =e^{g\circ g}, \\ g(0)=0,
\end{cases}$$
satisfied by $g(x):=-f^{-1}(-x)$ (Indeed, by the rule of the derivative of an inverse, $(f^{-1})'(x)={1\over f'(f^{-1}(x))}=e^{-f^{-1}(f^{-1}(x))}$ so that $g'(x) =e^{g(g(x))}$; see also Tom Copeland's previous answer [here][3].)
Indeed, assume by contradiction the formal power series solution $x+{1\over2}x^2+{1\over2}x^3+{2\over3}x^4+\&c.$ to the above equation has a positive radius of convergence. Then, it extends uniquely by analytic continuation to a maximally-defined analytic function, still denoted $g$ (that is, defined on the largest positive interval $[0,a)$, for some $0<a\le+\infty$).
Note that the Taylor series of $g$ at $0$ has non-negative coefficients. This follows immediately by induction, equating the coefficients of $g'$ and $e^{g\circ g}$; incidentally, this series is the EGF of the positive integer sequence [OEIS A214645][2], as also remarked [here][3]. As a consequence (check the details below), $g$ is totally monotonic on $[0,a)$; in particular $g'(x)>g'(0)=1$ and $g(x)>x$ for all $0<x<a$, and $g$ is invertible.
Then observe that $\log( g'( g^{-1}(x))$ is a well-defined analytic function on the interval $g[0,a)$, and coincides with $g$ locally at $0$. By the maximality of $[0,a)$ we have thus $g[0,a)\subset[0,a)$, but, due to the inequality $g(x)>x$ on $(0,a)$, this inclusion is only possible if $a=+\infty$, so that $g$ is unbounded. On the other hand, arguing as in Christian Remling's previous [answer][1], since $e^{-g(g(t))}g'(t)=1$ and $g(t)\ge t$, we have for any $x\ge0$
$$x=\int_0^{x}e^{-g(g(t))}g'(t)dt=\int_0^{g(x)}e^{-g(s)}ds\le \int_0^{+\infty}e^{-s}ds=1 ,$$
a contradiction.
$$*$$
**Rmk 1.** To justify the total monotonicity of $g$, note that, as a general elementary fact, *a real analytic function on an interval $I$, whose Taylor series at some point $x_0\in I$ has non-negative coefficients, has Taylor series with non-negative coefficients ay any point $x\in I$, $x\ge x_0$.* Indeed, this is clear for $x_1\ge x_0$ within the radius of convergence of $x_0$, and since there is a uniform radius of convergence at any $y\in [x_0,x]$, one reaches $x$ by finitely many steps $x_0<x_1<\dots<x_n=x$.
*(edit 5/29/21)* In fact more is true: *a real analytic function on $\mathbb R$, whose Taylor series at some point $x_0\in\mathbb R$ has non-negative coefficients is an entire function*, so that any $x\ge x_0$ is reached in just one step.
$$*$$
**Rmk 2.** The very same argument works for other differential-functional equations like e.g.
$$\begin{cases} g' =1 + {g\circ g}, \\ g(0)=0,
\end{cases}$$
that generates the sequence [OEIS A001028][4].
As before, a maximally-defined analytic solution $g$, if any, must be totally monotonic and defined for all $x\ge0$, for otherwise $ g'\circ g^{-1} -1$ would be a proper extension of it. Then we reach a contradiction as before, with one more step needed: since we have ${ g'(t)\over 1+g(g(t))}=1$ and $g(t)\ge t$ for any $t\ge0$, we also have, for any $x\ge0$
$$x=\int_0^{x}{ g'(t)dt\over 1+g(g(t))}=\int_0^{g(x)}{ dt\over 1+g(t)}\le\int_0^{g(x)}{ dt\over 1+t}=\log(1+g(x)) ,$$
whence $e^x\le 1+ g(x)$; if we plug this into the latter inequalities again, we get
$$x=\int_0^{g(x)}{ dt\over 1+g(t)}\le \int_0^{g(x)}e^{-t}dt\le 1 ,$$
as before. By comparison, the same conclusion also holds for $g'=F( {g\circ g})$ with any $F$ analytic and totally monotonic on $(-\epsilon,+\infty)$, and with $F(0)=1$.
[1]:https://mathoverflow.net/questions/258525/how-do-i-solve-this-displaystyle-f-ef-1/258544#258544
[2]:http://oeis.org/A214645
[3]:https://mathoverflow.net/questions/258525/how-do-i-solve-this-displaystyle-f-ef-1/258695#258695
[4]:http://oeis.org/A001028