Here's my attempt to construct a non-analytic solution. The idea is to use one of the analytic solutions in the accepted answer when x belongs to a set of points reachable through successive application of either analytic solution or their inverses, and the other solution otherwise. This set must not be equal the set of reals, otherwise the solution just reduces to one of the analytic solutions. I thought of a way to show this set is countable, and therefore can't be equal the real set, but realized it's easier to show by counterexample after finding the set.

I start with a couple of observations to facilitate finding this set. The first is the convenient commutativity of f1 and f2:

Simplify[f1[f2[x]] == f2[f1[x]]]

which means any sequence of f1 and f2 is equivalent to one with the same total number n1 and n2 of f1 and f2 calls respectively, e.g. n1 f1 calls followed by n2 f2 calls. This commutativity also implies the inverses are commutative, and sequences of InverseFunction[f1] and InverseFunction[f2] can be similarly ordered.

The second is the nth application of f1 or f2 is the sum of a constant term, which is a geometric series of length n, and a linear term, which is the nth term in another geometric series. These terms can be computed by hand, but Mathematica gives a convenient way to do so:

s1[x_]=FindSequenceFunction[NestList[f1,x,3]];
s2[x_]=FindSequenceFunction[NestList[f2,x,3]];

It can be shown the nth application of InverseFunction[f1] or InverseFunction[f2] to x is equal s1[x][-n] or s2[x][-n] respectively.

With these results, the set of points that can be reached from x0 satisfies:

criteria = Simplify[Reduce[{s2[s1[x][n1]][n2] == x0}, {n1, n2}], 
                    Assumptions -> {{n1, n2} \[Element] Integers, {x, x0} \[Element] Reals}]

$$ \left(x=-\frac{7}{2}\land \text{x0}=-\frac{7}{2}\right)\lor \left(c_1\in \mathbb{Z}\land 2 x+7\neq 0\land 2 \text{x0}+7\neq 0\land \\ \log (3) (\text{n1}+\text{n2})-2 i \pi c_1=\log \left(-\frac{9 (-1)^{\text{n1}} (2 \text{x0}+7)}{2 x+7}\right)\right) $$

Unless x0 is the fixed point -7/2, we can reach x if there are integers n1 and n2 satisfying this equation.

Here, the only imaginary terms are $2i\pi c_1$ and from the log on the right-hand side. Since x and x0 are real, the imaginary part of the log can only be 0 or $i\pi$. The former requires $c_1=0$, while the latter cannot be satisfied for any $c_1\in \mathbb{Z}$. We can therefore set $c_1=0$.

Moreover, given x and x0, there are only two possibilities for the log, one with n1 even, and one with n1 odd. Regardless of whether n1 is even or odd, n2 is arbitrary, so n1+n2 need only be integral and can be either even or odd:

condition[x_,x0_]=And @@ Reverse @ (Or @@ (n\[Element]Integers/.Simplify[
          Solve[criteria/.n1+n2->n,n][[1]],Assumptions->{x,x0}\[Element]Reals]/.C[1]->0/.#& 
          /@ {n1->1,n1->2}))

$$ x\neq -\frac{7}{2}\land \text{x0}\neq -\frac{7}{2}\land \left(\frac{\log \left(-\frac{9 (2 \text{x0}+7)}{2 x+7}\right)}{\log (3)}\in \mathbb{Z}\lor \frac{\log \left(\frac{9 (2 \text{x0}+7)}{2 x+7}\right)}{\log (3)}\in \mathbb{Z}\right) $$

This condition is sufficient to show the set of points that can be reached is not the real set, e.g. condition[1,0] returns False, showing x0=0 cannot be reached from x=1.

With this condition, we construct a non-analytic solution, as f1[x] for x satisfying the condition, and f2[x] otherwise:

f[x_, x0_] = Piecewise[{{f1[x], condition[x, x0]}}, f2[x]];

Testing:

ff=FullSimplify[f[f[x,x0],x0],Assumptions->{x,x0}\[Element]Reals]

unfortunately gives a long condition for 28+9x I have to simplify by brute force:

(ff[[1,1,2]]/.Log[x_]:>Log[Factor[x]]//PowerExpand//ExpandAll//FullSimplify)/.
(Log[x_]-Log[y_]:>Log[x/y])/.I \[Pi]+Log[x_]->Log[-x]/.
{x_\[Element]Integers->integerQ[x],x_\[NotElement]Integers:>!integerQ[x]}//FullSimplify

True