Suppose that $f(t)=a(t)g(t)+b(t)$ for $t\geq0$, where $a$ and $b$ are continuous functions. Thus, once can immediately can say that if $b\in{}L^{p}$, $\liminf_{t\to\infty}|a(t)|>0$ and $\limsup_{t\to\infty}|a(t)|<\infty$, then $f\in{}L^{p}$ if and only if $g\in{}L^{p}$.
Question. What can be the other possible natural conditions on $a$ and $b$ under which $f\in{}L^{p}\iff{}g\in{}L^{p}$?
Thanks for your clarifications. As far as I can see, there is no way to really weaken those conditions, although the one for $a$ might be reformulated.
But first of all, if you put $g=0$, then $f=b$, so we have to have $b\in L^p$ as a condition no matter what.
The condition on $a$ boils down to whether $a$ is bounded both from above as well as away from 0. So they should be $a\in L^\infty$ and $1/a\in L^\infty$, I'm sure this is what you meant by your condition. However, those are not avoidable as well.
Assume that $a$ is not in $L^\infty$. Then the sets $A_k :=\{k+1 \geq a(x)> k\}$ have nonzero measure for all $k \in \mathbb{N}$. We can assume that they are not of infinite measure as well (Otherwise cut them of by using $\tilde{A}_k = A_k \cap [0,C)$ for some $C$ depending on $k$, such that $|\tilde{A}_k|$ is still nonzero).
Then define $g$ by $g(x):= \left(\frac{1}{|A_k|k^{1+\epsilon}}\right)^{1/p}$ for $x \in A_k$, respectively $g(x)=0$ if $x$ is in none of the $A_k$. Now with a bit of calculation you get that $g \in L^p$, since $\|g\|_{L^p} = \sum_{k=1}^\infty \frac{1}{k^{1+\epsilon}}$. However by a similiar calculation you get $a \cdot g \notin L^p$ for $\epsilon$ small enough, so $f\notin L^p$.
This also works for $1/a \notin L^\infty$, since then $g = 1/a \cdot f - b$ which is the same situation.
So, tl,dr: The set of all possible pairs $a,b$ that satisfy your condition is characterized, by $b\in L^p$, $a,1/a \in L^\infty$, which is essentially what you already have minus continuity.