When we attempt a transformation to a new variable $x\rightarrow y$ we have to conserve the probability $$\begin{equation*} p(y)dy=p(x)dx, \end{equation*}$$ which for the uniform distribution implies $$\begin{equation*} p(y)dy=dx. \end{equation*}$$ Let us assume that $p(y)$ is a PDF different from the uniform PDF $p(x)=1$ with $x \in [0,1]$. If we integrate the last expression we arrive at $$\begin{equation*} x(y)=\int_0^y p(y')dy', \end{equation*}$$ which is nothing but the cumulative distribution of $p(y)$, i.e., $$\begin{equation*} x(y)=P(y)=\int_0^y p(y')dy'. \end{equation*}$$