The assumption $2\pi xy=t$ (or something similar) is certainly necessary to guarantee that you get an approximation to the zeta-function and not some other function. For instance, if we choose $x=y=|t|$ then the approximate functional equation "approximates" twice the zeta-function. One can see this as follows:

Inside the critical strip ($\frac{1}{4}\le\sigma\le \frac{3}{4}$, say), it is known that $$ \zeta(s) = \sum_{n\leq |t|} \frac{1}{n^s} +O(|t|^{-\sigma})$$ for sufficiently large $|t|$. Hence, by the functional equation and Stirling's formula, we have $$ \zeta(s) =\chi(s)\zeta(1-s) = \chi(s)\sum_{n\leq |t|} \frac{1}{n^{1-s}} +O(|t|^{1/2}),$$ as well. Therefore, $$ 2\zeta(s) = \sum_{n\leq |t|} \frac{1}{n^s} + \chi(s)\sum_{n\leq |t|} \frac{1}{n^{1-s}} +O(|t|^{-\sigma}) +O(|t|^{1/2}),$$ as claimed.

The assumption $2\pi xy=t$ (or something similar) is certainly necessary to guarantee that you get an approximation to the zeta-function and not some other function. For instance, if we choose $x=y=t/2\pi$ then the approximate functional equation "approximates" twice the zeta-function. One can see this as follows:

Inside the critical strip ($\frac{1}{4}\le\sigma\le \frac{3}{4}$, say), it is known that $$ \zeta(s) = \sum_{n\leq \frac{t}{2\pi}} \frac{1}{n^s} +O(t^{-\sigma})$$ for sufficiently large $t.$ Hence, by the functional equation and Stirling's formula, we have $$ \zeta(s) =\chi(s)\zeta(1-s) = \chi(s)\sum_{n\leq \frac{t}{2\pi}} \frac{1}{n^{1-s}} +O(t^{-1/2}),$$ as well. Therefore, $$ 2\zeta(s) = \sum_{n\leq \frac{t}{2\pi}} \frac{1}{n^s} + \chi(s)\sum_{n\leq \frac{t}{2\pi}} \frac{1}{n^{1-s}} +O(t^{-\sigma}) +O(t^{-1/2}),$$ as claimed.

Edit: You can also see this choosing $x=\frac{t}{2\pi}$ and $y=1$ (and vice versa) in formula that you stated in your question.