Indeed, you are right. If we denote with $\mathbb{P}(A)$ the probability of the event $A$ to happen and, use the fact that $xy$ is even if and only if either $x$ or $y$ is, we get 
$$
\mathbb{P}(\{xy \; \text{is even}\}) = \mathbb{P}(\{x \; \text{is even}\} \cup \{ y \; \text{is even}\})
$$
Finally, we can split the event in all the three distinct cases, i.e. $x$ even, $y$ even or $x$ and $y$ are both even. Concluding that 
$$
\mathbb{P}(\{x \; \text{is even}\} \cup \{ y \; \text{is even}\})
= \mathbb{P}(\{x \; \text{is even}\} \cup \{ y \; \text{isn't even}\}) + \mathbb{P}(\{x \; \text{isn't even}\} \cup \{ y \; \text{is even}\}) + \mathbb{P}(\{x \; \text{is even}\} \cup \{ y \; \text{is even}\})
$$ 
because all of the above events have same probability $1/4$ we get the aforementioned claim.
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