As @dxiv mentioned, yes, it needn't happen always. I'd try and say using @dxiv's example:
We know, $\gcd(3,2) = 1 = d$. By Bezout's identity, we can say that $\exists s,t : 3s+2t = 1$.Here we see a solution $s = 1, t = -1$. Suppose you take $s$ as an even integer, say $2n$, we see $6n + 2t = 1$ which has no solution in $\mathbb{Z}$ as no two even numbers differ by $1$. But taking an even $t$ may be possible, for example let's say $t = 2k$ and thus $3s + 4k = 1$ and we get $(s,k) = (-9,7)$ and hence $(s,t) = (-9, 14)$.So yes, if you ever get an even $s'$ or $t'$, it's just trivial or even more properly, is a result of the numbers used(perhaps; I'm still learning 😄). If you look up in Wikipedia about Bezout's Identity, you can see the example of $12$ and $42$; here $s'$ can be even. So an even $s'$ can appear trivially or it may not; all depends on the numbers used.