I think a conceptual manner of understanding this situation is to define a positive measure $\mu$ on $\mathbb{R}^n$ by the rule $d\mu = \left|g\right|dy$, where $dy$ is Lebesgue measure, i.e., by the rule $\mu(E)=\int_{E} \left|g(y)\right| dy$, where $E$ is a Lebesgue measurable subset of $\mathbb{R}^n$. (In the language of measure theory, $\left|g\right|$ is the Radon-Nikodym derivative $\frac{d\mu}{dy}$.)
If $x\in\mathbb{R}^n$ is fixed, then let us apply Hölder's inequality to the functions $y\to \left|f(x-y)\right|$ and $y\to 1$ on $\mathbb{R}^n$ with respect to the measure $\mu$ and with the exponents $p$ and p':
\begin{eqnarray*} \int_{\mathbb{R}^n} \left|f(x-y)g(y)\right| dy &=& \int_{\mathbb{R}^n} \left|f(x-y)\right|d\mu(y) \\ &\leq& \left(\int_{\mathbb{R}^n} \left|f(x-y)\right|^{p}d\mu(y)\right)^{\frac{1}{p}}\left(\int_{\mathbb{R}^n} 1^{p'}d\mu(y)\right)^{\frac{1}{p'}} \\ &=& \left(\int_{\mathbb{R}^n} \left|f(x-y)\right|^{p}\left|g(y)\right|dy\right)^{\frac{1}{p}}\left(\int_{\mathbb{R}^n}\left|g(y)\right|dy\right)^{\frac{1}{p'}}. \end{eqnarray*}
You should justify the equalities in the computation above (the inequality is simply a special case of Hölder's inequality). If we take the $p$th power of both sides of the inequality in the above computation, then we deduce the desired inequality.
The technique that I have used to answer your question is an important one; e.g., it is used in at least one proof of Young's inequality on convolutions (if I remember correctly).
I hope this helps!