This works, right?
If we let $z$ = $(p_0, p_1, p_2)$ and $\zeta$ = $(q_0, q_1, q_2)$, and if we say $z \diamond \zeta$ defines the operation $(\frac{p_0}{q_0}, \frac{p_1}{q_1}, \frac{p_2}{q_2})$, then if each $\frac{p_i}{q_i} $ is equal for all $i$ = 0, 1, 2, then $x$ is parallel to $y$.
The definition provided in the book is different than the definition you have created. How can we show this? Simply provide a counter-example:
Let $\vec a = \langle 1, 0, 4 \rangle$, and $\vec b = \langle 2, 0, 8 \rangle$.
These vectors are parallel by the book's definition, as $\vec a = 2\vec b$.
However, using your $\diamond$ operator: $$\vec a \diamond \vec b = \left\langle \frac{1}{2}, \frac{0}{0}, \frac{4}{8}\right\rangle$$
We cannot conclude that $\frac{1}{2}= \frac{0}{0}= \frac{4}{8}$, as $\frac{0}{0}$ is undefined.
Thus, your definition is not equivalent.