This is too long for a comment.

I don't think the minimum always exists, i.e. that Bézout's identity can be fulfilled. In $\mathbb Q[x,x^{-1}]$ it would be the case, but not in $\mathbb Z[x,x^{-1}]$. Take $P=x+\frac1x$ and $Q=-2$. Supposing there are $A,B\in\mathbb Z[x,x^{-1}]$ such that $AP+BQ=1$, say

$A=a_nx^n+\cdots+a_{-n}x^{-n}$ and $B=b_nx^n+\cdots+b_{-n}x^{-n}$ (possible by choosing $n$ big enough). Equating coefficients we get the system

$a_n\qquad \quad =0$

$a_{n-1}\qquad =2b_n$

$a_{n-2}+a_n=2b_{n-1}$

$...$

$a_0+a_2=2b_1$

$a_{-1}+a_1=2b_{0}+1$

$a_{ -2}+a_0=2b_{-1}$

$...$

$a_{-n}+a_{-n+2} =2b_{-n+1}$

$.\qquad a_{-n+1} =2b_{-n}$

$.\qquad a_{-n}\quad =0$

So we find successively from the top $a_{n-1}\equiv a_{n-2}\equiv \cdots\equiv a_1\equiv 0\pmod 2$ and from the bottom $a_{-n+1}\equiv a_{-n+2}\equiv \cdots\equiv a_{-1}\equiv 0\pmod 2$ which yields a contradiction in the middle.

So even though the $gcd$ may be well-defined, it looks like $\mathbb Z[x,x^{-1}]$ is not an Euclidean domain.