Let $p(y)$ denote the probability distribution function with a Poisson random variable with mean $λ$.
I've proven in another part of this question that $p(y)\over p(y-1)$ = $λ \over y$ for all $y = 1, 2, ...$ .
I'm asked: consider which values of $y$ have $p(y) > p(y-1)$ and use it to show that $p(y)$ is maximized when $y = $ the greatest integer less than or equal to $λ$.
Now I believe that $p(y) > p(y-1)$ whenever $ λ > y$. I think what I need to show next is that as $y$ increases, $p(y)$ continues to increase until a certain point, at which $p(y)$ then begins to decrease. This turning point will happen to be where $y = $ the greatest integer less than or equal to $λ$. But I am having trouble figuring out how to show this?
You have got the idea. Note that the pmf $p$ is positive, so as you said we have $$\begin{cases} p(y) > p(y - 1) \text{ when } y < \lambda \\ p(y) < p(y-1) \text{ when } y > \lambda \end{cases}$$
So when $\lambda$ is not an integer, we have $p(1) < p(2) < \ldots < p(\lfloor \lambda \rfloor)$ and $p(\lfloor \lambda \rfloor) > p(\lceil \lambda \rceil) > p(\lceil \lambda \rceil + 1) > \ldots$. So $\lfloor \lambda \rfloor$ is the maximum point.
When $\lambda$ is an integer, we have $p(\lambda) = p(\lambda - 1)$ and both of them are maximum. Of course in this case $\lfloor \lambda \rfloor = \lambda$ also so you have shown the thing you need.