No, this is not true. Here is an indirect argument (if I made no mistake).

If the statement would be true for $\epsilon > 0$ small enough, then it would be true for all $\epsilon > 0$ (since the maps commute, the set of "good" epsilons for a given $(a,b,c)$ is open; this set is also closed because the Lipschitz condition is closed). We will construct two triangles $(a,b,c)$ and $(a+M, b+M, c+M)$ for which the latter does not dominate the former.

Take $a=b=1$ and $c=2$ (or slightly smaller if you need). Take points $A_1$ and $B_1$ on the sides $BC$ and $AC$ at distance $\frac13$ from $C$. The distance between $A_1$ and $B_1$ is $\frac23$. Now consider a triangle $(x, x, x+1)$ for a very large $x$. I claim that the distance between the corresponding points $A'_1$ and $B'_1$ tends to $0$ as $x$ tends to $\infty$.

By the sine law in a right triangle we have
$$\frac{\sinh \frac{A'_1B'_1}{2}}{\sinh\frac{x}{3}} = \sin\frac{\gamma'}{2} = \frac{\sinh\frac{x+1}2}{\sinh x}$$
which implies that $\sinh\frac{A'_1B'_1}2$ goes down as $e^{-\frac{x}6}$.

This means that somewhere inbetween there is an isosceles triangle for which the map is not $1$-Lipschitz even for small $\epsilon$...