No such class $F$ exists, and in fact we do not even need condition 3.

***

First of all, note that every function $f$ in $F$ is non-decreasing and $f(1) = 1$.

Consider the composition $g = f_n \circ \ldots \circ f_1$. For every $x \in (0, 1)$, unless $g(x) = 1$, we have
$$ \log g'(x) = \sum_{i = 1}^n \log f_i'(x_i) , $$
where $x_i = f_{i-1} \circ \ldots \circ f_1(x)$. Therefore,
$$ (\log g'(x))' = \sum_{i = 1}^n (\log f_i'(x_i))' = \sum_{i = 1}^n \frac{f_i''(x_i)}{f_i'(x_i)} \geqslant \frac{n m}{M} \geqslant \frac{m}{M} \, . $$
It follows that if $a < x$ and $g(x) < 1$, then
$$ \log \frac{g'(x)}{g'(a)} \geqslant \frac{m (x - a)}{M} \, ,$$
and thus
$$ \frac{g(b) - g(a)}{g'(a)} \geqslant \int_a^b \exp\biggl(\frac{m (x - a)}{M}\biggr) dx \tag{1} $$
whenever $a < b$ and $g(b) < 1$. Similarly, if $x < b$ and $g(b) < 1$, then
$$ \log \frac{g'(x)}{g'(b)} \leqslant -\frac{m (b - x)}{M} \, ,$$
and consequently
$$ \frac{g(b) - g(a)}{g'(b)} \leqslant \int_a^b \exp\biggl(-\frac{m (b - x)}{M}\biggr) dx \tag{2} $$
whenever $a < b$ and $g(b) < 1$.

We proceed by contradiction. Fix a sufficiently small $\varepsilon \in (0, \tfrac13)$ (to be specified later), and suppose that $g$ is $\varepsilon$-close to the identity function. Note that $g(\tfrac23) \leqslant \tfrac23 + \varepsilon < 1$. Set $a = \tfrac13$ and $b = \tfrac23$ in (1), and $a = 0$ and $b = \tfrac13$ in (2) to find that
$$ \frac{g(\tfrac23) - g(\tfrac13)}{g'(\tfrac13)} \geqslant \int_{\frac13}^{\frac23} \exp\biggl(\frac{m (x - \tfrac13)}{M}\biggr) dx =: A $$
and
$$ \frac{g(\tfrac13) - g(0)}{g'(\tfrac13)} \leqslant \int_0^{\frac13} \exp\biggl(-\frac{m (\tfrac13 - x)}{M}\biggr) dx =: B , $$
where $A > 1$ and $B < 1$ depend only on $m$ and $M$. It follows that
$$ \frac{g(\tfrac13) - g(0)}{g(\tfrac23) - g(\tfrac13)} \leqslant \frac{B}{A} \, , $$
which, in turn, implies that
$$ \frac{\tfrac13 - 2 \varepsilon}{\tfrac13 + 2 \varepsilon} \leqslant \frac{B}{A} \, . $$
If, however, $\varepsilon$ was chosen sufficiently small, the above inequality is false, and the proof is complete.