identifying the measure $\lambda f^{-1}$ on the interval $[0,1]$

If you can prove the statement for all open dyadic intervals it would be already very useful. (I assume that by dyadic interval you mean an interval whose endpoints are of the type $k2^{-n}$ for suitable integers $k,n$.)

If you have $\lambda f^{-1}(E)=m(E)$ for open dyadic intervals $E$, then one can show that $\lambda f^{-1}(E)=m(E)$ also holds for all (non-dyadic or dyadic) open intervals $E\subset [0,1]$. This can be done by approximating the open intervals by dyadic intervals from inside: If you have real numbers $a,b\in [0,1]$ with $a<b$, then there exist sequences $k_n,l_n\in\Bbb N$ such that $x_n:= k_n2^{-n}$ converges from above to $x$ and $y_n:=l_n2^{-n}$ converges from below to $b$. For large $n$, the sequences $k_n,l_n$ can be chosen such that $a\leq x_n\leq a+2^{-n} < b-2^{-n} \leq y_n \leq b$ is satisfied. Since the interval $(x_n,y_n)$ is a dyadic interval, we have $\lambda f^{-1}((x_n,y_n))=\mu((x_n,y_n))=y_n-x_n$. Using the properties of a measure (like continuity from below) it follows that $\lambda f^{-1}((a,b))=\mu((a,b))=y_n-x_n$ holds for all real numbers $a,b\in [0,1]$.

If two measures are equal on all open intervals, then it is known that these measures agree on all Borel measurable sets, see for example this question and its comments and answers (the fact that you use $[0,1]$ while the question uses $\mathbb R$ does not make a significant difference, the arguments work the same in both cases).

Thus we can conclude that $\lambda f^{-1}$ is just the standard Borel measure on $[0,1]$.