Is the following a PRG?
I believe that it is not, but I could easily be wrong.
... I'm having trouble reaching the conclusion of my contradiction

The problem is you can't show that contradiction, because there isn't any. As a rule of thumb for homework questions: If it becomes clear that you can't prove your initial assumption, consider that your intuition was wrong.
Here are some pointers how you can show that $G'$ is in fact a PRG, in a proof by contradiction:

Let's assume $G'$ is not a PRG, and that means a distinguisher $\mathcal{D}$ exists.
The input for $\mathcal{D}$ is one element either uniform random or of the form $x||(G(y)\oplus x)$, and then it has some non-negligible advantage $\epsilon$.
We want to build a distinguisher for $G$. So as input we get one element, which is either from a uniform random distribution or from the image of $G$.
In general, if $a$ is uniform random, then $a \oplus b$ is also uniform random -  this is true as long as $b$ is independent of $a$. This includes fixed $b$ as well as $b$ drawn independently from any distribution.