Here $a(t,M)$, $b(t,M)$, $c(t,M)$, $x(t,M)$, and $y(t,M)$ are Laurent polynomials in $t$ and $M$ with rational coefficients.
I am given $a(t,M)$, $b(t,M)$, and $c(t,M)$. Is there a way for me to tell if a solution in $x(t,M)$ and $y(t,M)$ there exists for the following equation $$a(t,M)x(t,M)+b(t,M)y(t,M)=c(t,M).$$
If this solution exist, can I say that $\gcd(a(t,M),b(t,M))$ divides $c(t,M)$?
If such a solution exists, then it is a multiple of the greatest common divisor (because a sum of multiples of an element is a multiple of that element).
For your first question, I believe you are looking for something such as Bézout's identity: if we were in a Bézout domain, we could say that such a solution exists if the RHS is a multiple of the greatest common divisor.
But I don't think your ring is a Bézout domain. The ring $\mathbb{Q}[t]$ is a PID, and so is its localization $R=\mathbb{Q}[t,t^{-1}]$. Hence $R$ is a Bézout domain. But now, $R[M]$ is not a PID any more, neither is its localization $S=R[M,M^{-1}]$. To show now that your ring $S$ is not a Bézout domain, try to argue like you would in $\mathbb{Q}[x,y]$: the $gcd$ of $x$ and $y$ is $1$, but $1\notin (x,y)$.