Let $f$ be a function, for example $f(x)=log(1+x)$ and let $c$ be some constant $>0$ (for simplicity, we may assume that it is different from 1).
Is this true:
$$\frac{f(x)}{1-c-o(1)}= \frac{f(x)}{1-c}(1-o(1))?$$
What is the difference between $-o(1)$ and $+o(1)$, when both contain positive and negative functions?
You may observe that, as $u \to 0$, you have $$ \frac1{1-u}=1+u+O(u^2) $$ implying $$ \frac1{1-u}=1+O(u) \tag1 $$ as $u \to 0$.
Here, using $(1)$, you may write, for $c \neq 1$, $$ \frac{f(x)}{1-c-o(1)}=\frac{f(x)}{1-c}\left(\frac1{1- o(1)}\right)=\frac{f(x)}{1-c}\left(1+ o(1)\right) $$ since clearly $\dfrac{o(1)}{1-c}=o(1)$ and $O(o(1))=o(1)$.