**Non-autonomous case**

Most of the time, either the few usual tricks work, or the conjecture is true. In the non-autonomous case, unfortunately, the answer is the former. This is the first counterexample I ever learnt to uniqueness (and I bet it's true for a lot of people):

$$
x' = 2 \max(x,0)^{1/2}
$$

That has two solutions, if $x(0) = 0$. One is $x(t) = 0$ and one $x(t) = t^2\cdot 1_{t>0}$. Because you want to assume boundedness (as you will see it's not really important, let's consider the bounded counterpart

$$
x' = 2\cdot
\begin{cases}
0 & \text{if }t\le 0 \\
\sqrt{x} & \text{if } t\in [0,1]\\
1 & \text{if }t\ge 1 \\
\end{cases}
$$

Let $x(t)$ be a solution to the problem above, and choose an $v>0$. What equation does $z(t) = x(t)+vt$ solve? A direct computation shows:

$$
z' = 2\cdot
\begin{cases}
v & \text{if }z-vt\le 0 \\
v + \sqrt{z -vt} & \text{if } z-vt\in [0,1]\\
1+v & \text{if }z-vt\ge 1 \\
\end{cases}
$$

So you have a counterexample for nonautonomous solutions. From a philosophical perspective this boils down to the following:

In the non-autonomous case, you have time and space reparametrization symmetries. (At least you reparametrize by smooth changes of coordinates). If you want to show that a statement holds for a certain class of objects, you should look for a class that is invariant under reparametrizations.

**Autonomous case**

Now, in the autonomous case, the class you have given *is* invariant by the (reduced set of) symmetries of your problem, at least locally, and that gives hope. (I found an easier solution that shows uniqueness assuming existence). Assume that a solution $x'(t)=F\circ x(t)$ exists (in the mild/integral sense). Then $t\mapsto x(t)$ must be a bi-liptschitz map by the hypohtesis, and therefore must have a (bi-liptschitz inverse) $x \mapsto t(x)$. By the inverse function theorem, it must be that

$$
t(x) = \int_0^x 1/F(x) dx
$$

assuming wlog that $x(0)=0$. This tells you who $t(x)$ is, so uniqueness is set. You might, however, ask for well-posedeness of the solution. This will be a bit harder, but we'll also get an existence result from it (by density). Define three spaces:

$$
\begin{split}
\mathcal F &= (\{f\in L^1([-M,M]), f(x) \in [1/M,1]\}, \|\cdot\|_{L^1})\\
\mathcal T &= (\{f\in L^{1,1}([-M,M]), f(x) \in [-M^2,M^2], f(0) = 0, f'(x) \in [1/M,M]\}, \|\cdot\|_{W^{1,1}})\\
\mathcal X &= (\{f\in L^{1,1}([-1,1]), f(x) \in [-M,M], f(0) = 0, f'(x) \in [1,M]\}, \|\cdot\|_{L^{\infty}})\\
\end{split}
$$
As the names already tell, the first is the space where $F$ lives (it will actually be where $1/F$ lives), the second where $x\mapsto t(x)$ lives, and the third one where $x\mapsto x(t)$ lives.

We will define two continuous maps, $\int:\mathcal F \to \mathcal T$ the indefinite integral. It is continuous by construction of the spaces. The magic is that there is a unique continuous map $I:\mathcal T \to \mathcal X$ defined implicitly by $\tau(I(\tau)(t)) = t$. In other words $x = I(\tau)$ is the left inverse of $\tau$, and therefore a solution to the equation by the inverse function theorem.

Let's now show that the map $I$ is continuous. Let $\tau, \tau'\in \mathcal T$. Then $\chi = \tau^{-1}, \chi' = {\tau'}^{-1}$ exist and are Lipschitz. Let $(x,t)$ be a point in $(x,\tau(x))$. By the mean value theorem for bi-Lipschitz functions applied to $\tau'$ at the points $x,\chi'(x)$, we see $$
\frac{|t-\tau'(x)|}{|x-\chi'(t)|} \in [1/M,1]
$$ (Here we use the hypothesis!). In particular
$$
\|\chi-\chi'\|_\infty < M\|\tau-\tau'\|_\infty.
$$

We have then shown that the data-to-solution map is well defined and bi-Liptschtiz in the given metrics. However, we have "lost" a derivative in the process. We should be able to do better. (I think the map $I$ is not continuous if you endow $\mathcal X$ with the $W^{1,1}$ topology, so this proof won't work)

The "Riemann Sum" solution that I was proposing before is a discrete version of this proof, which gets messier because you're discretizing everything, and weaker.

**General existence**

You might be able to get existence under very mild assumptions using Schauder's fixed point, which doesn't ask for a contraction.