Yes, there is such a preorder. I will argue that there is a preorder on the space of bounded functions $\mathbb Z\to\mathbb R$ so that comparing indicator functions in this space does the job. A vector space preorder can be constructed from a suitable "positive cone", the set of non-negative elements, so the main task is to construct this cone.

Let $M$ be the set of non-negative, not identically zero, finitely-supported functions $\mathbb Z\to\mathbb R.$ Let $B$ be the real vector space of bounded functions $\mathbb Z\to\mathbb R.$ Let $a*\phi$ denote convolution of a function $a\in B$ by a function $\phi\in M.$ Define $a\sim b$ for $a,b\in B$ to mean that $a*\phi=b*\psi$ for some $\phi,\psi\in M.$ This is an equivalence relation because $M$ is a (commutative) monoid under convolution. The $\sim$-equivalence class $[0]$ of zero is a linear subspace of $B.$

Define a good cone to be a set $C\subset B$ such that

- C1. $y\in C\iff z\in C$ whenever $y\sim z,$ and
- C2. $C$ is a convex cone ($x,y\in C\implies \lambda x+\mu y\in C$ for $\lambda,\mu\geq 0$), and
- C3. $C\cap (-C)=[0].$

Define $C_0$ to be the set of $x\in B$ such that $x\sim y$ for some non-negative function $y\in B.$ Because non-negative functions are closed under convolution by any $\phi\in M,$ the definition of $C_0$ simplifies slightly to $x*\phi$ being non-negative for some $\phi\in M.$ The set $C_0$ satisfies the good cone conditions: (C1) is obvious, for (C2) if $x*\phi$ and $y*\psi$ are non-negative and $\lambda,\mu\geq 0$ then $(\lambda x+\mu y)*\psi*\psi$ is non-negative, and for (C3) if $x*\phi$ is non-negative and $x*\psi$ is non-positive, then $x*\phi*\psi$ is identically zero so $x\sim 0.$ By Zorn's lemma there is a maximal good cone $C$ containing $C_0.$

Consider $x\in B\setminus C.$ Define $C_x$ to be the set of $y\in B$ such that $y*\phi= x*\psi+c$ for some $\phi\in M$ and $\psi\in M\cup\{0\}$ and $c\in C.$ By maximality of $C,$ the set $C_x$ is not good.
(C1) holds: whenever $y*\eta=z*\zeta$ and $y*\phi= x*\psi+c$ we have $z*\zeta*\phi=x*\psi*\eta+c*\eta,$ which implies $z\in C_x.$
(C2) holds: if $y*\phi= x*\psi+c$ and $y'*\phi'= x'*\psi'+c'$ and $\lambda,\mu\geq 0$ then $(\lambda y+\mu y')*\phi*\phi'=x*(\psi*\phi'+\psi'*\phi)+(c*\phi'+c'*\phi).$
So (C3) fails: some $y\not\sim 0$ satisfies $y*\phi= x*\psi+c$ and $y*\phi'=-x*\psi'-c'.$ But then $$-x*\psi'*\phi-c'*\phi=y*\phi*\phi'=x*\psi*\phi'+c*\phi'$$
which implies $x*(\psi*\phi'+\psi'*\phi)+(c*\phi'+c'*\phi)=0.$ If $\psi$ and $\psi'$ are both zero, then $c*\phi=-c'*\phi$ is in $C\cap (-C)$ contradicting $y\not\sim 0.$ So $\psi$ and $\psi'$ are not both zero, which means $-x\sim c*\phi'+c'*\phi\in C.$
In other words $x\in -C.$

We have shown that $C\cup (-C)=B.$ The cone $C$ defines a total vector order on $B/[0],$ but to answer the question we just need to define $S\lesssim T\iff 1_T-1_S\in C.$ Your condition 1 comes from $C_0\subset C$ and (C3). Your condition 2 comes from (C1) - shifting is convolution by a delta function. Your condition 3 comes from $1_{T\cup U}-1_{S\cup U}=1_T-1_S$ whenever $S\cap U=T\cap U=\emptyset.$

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