Identity for Landau symbols (little $o$) I was wondering if the following is true:

If $f=g+o(x)$, then $o(f)=o(g)$.

Here, $f=f(x)$, $g=g(x)$.
No, this is not true. 
With the remark that it's usually a very good idea to specify where the asymptotics are taken (i.e., the point $a\in [-\infty,\infty]$).

Now, take $f,g$ defined on $\mathbb{R}$ by $f(x)=x^2$ and $g(x)=x^4$. Consider the asymptotics at $0$: certainly,
$$
f(x) = g(x) + o(x) 
$$
as desired. Yet, we do have $x^3 = o(f(x))$ (at $0$, but definitely not $x^3=o(g(x))$.
The writing ''$o(f)=o(g)$'' does not make any sense: the symbol "little o" is not defined like an operator on functions, but only as $f(x)=o(g(x))$ for $x\rightarrow x_0$, meaning that the quotient of $f$ by $g$ has limit zero as $x\rightarrow x_0$, and that is it. So your equality does not make any sense.

To be more precise, and to clarify something which is not completely clear in the book, something that people often overlook, or simply do not know at all is that, from a logical point of view, when we define $f(x)=o(g(x))$in Mathematics what is defined it is not really the symbol ''$o$'', but the symbol ''$=o$'', and this is the reason why equations of the kind you wrote simply do not make sense.

The same thing is true for all the other ''order of growth'' symbols like $O$, $\sim$, etc. What we define are not these symbols per se, but the notions ''$=O $ '', ''$=\Theta$'', and so on.