$\let\sset\subseteq\DeclareMathOperator\span{span}$Using some ideas from the other answers, here is an argument which works for general $n\times n$ Sudokus, $n\ge 2$.
I will denote the cells as $s_{ijkl}$ with $0\le i,j,k,l< n$, where $i$ identifies the band, $j$ the stack, $k$ the row within band $i$, and $l$ the column within stack $j$. Rows, columns, and blocks are denoted $r_{ik},c_{jl},b_{ij}$ accordingly. Let $X=\{r_{ik},c_{jl},b_{ij}:i,j,k,l< n\}$ be the set of all $3n^2$ checks. For $S\sset X$ and $x\in X$, I will again denote by $S\models x$ the consequence relation “every Sudoku grid satisfying all checks from $S$ also satisfies $x$”.
Let $V$ be the $\mathbb Q$-linear space with basis $X$, and $V_0$ be the span of the vectors $\sum_kr_{ik}-\sum_jb_{ij}$ for $i< n$, and $\sum_lr_{jl}-\sum_ib_{ij}$ for $j< n$.
**Lemma 1:** If $x\in\span(S\cup V_0)$, then $S\models x$.
**Proof:** A grid $G$ induces a linear mapping $\phi_G$ from $V$ into the space $W$ with basis $e_1,\dots,e_{n^2}$ such that for any $x'\in X$, the $i$th coordinate of $\phi_G(x')$ gives the number of occurrences of the number $i$ in $x'$. We have $\phi_G(V_0)=0$, and $G$ satisfies $x'$ iff $\phi_G(x')$ is the constant vector $\vec 1$. If $x=\sum_i\alpha_ix_i+y$, where $x_i\in S$ and $y\in V_0$, then $\phi_G(x)=\vec\alpha$ for $\alpha:=\sum_i\alpha_i$. The same holds for every grid $G'$ satisfying $S$; in particular, it holds for any valid grid, which has $\phi_{G'}(x)=\vec1$, hence $\alpha=1$. QED
We intend to prove that the converse holds as well, so assume that $x\notin\span(S\cup V_0)$. We may assume WLOG $x=r_{00}$ or $x=b_{00}$, and we may also assume that $r_{i0}\notin S$ whenever $r_{ik}\notin S$ for some $k$, and $c_{j0}\notin S$ whenever $c_{jl}\notin S$ for some $l$. By assumption, there exists a linear function $\psi\colon V\to\mathbb Q$ such that $\psi(S\cup V_0)=0$, and $\psi(x)\ne0$. The space of all linear functions on $V$ vanishing on $V_0$ has dimension $3n^2-2n$, and one checks easily that the following functions form its basis:
* $\omega_{ik}$ for $0\le i< n$, $0< k< n$: $\omega_{ik}(r_{ik})=1$, $\omega_{ik}(r_{i0})=-1$.
* $\eta_{jl}$ for $0\le j< n$, $0< l< n$: $\eta_{jl}(c_{jl})=1$, $\eta_{jl}(c_{j0})=-1$.
* $\xi_{ij}$ for $i,j< n$: $\xi_{ij}(r_{i0})=\xi_{ij}(c_{j0})=\xi_{ij}(b_{ij})=1$.
(The functions are zero on basis elements not shown above.) We can thus write
$$\psi=\sum_{ik}u_{ik}\omega_{ik}+\sum_{jl}v_{jl}\eta_{jl}+\sum_{ij}z_{ij}\xi_{ij}.$$
If $r_{ik}\in S$, $k\ne0$, then $0=\psi(r_{ik})=u_{ik}$, and similarly $c_{jl}\in S$ for $l\ne0$ implies $v_{jl}=0$. Thus, the functions $\omega_{ik}$ and $\eta_{jl}$ that appear in $\psi$ with a nonzero coefficient individually vanish on $S$. The only case when they can be nonzero on $x$ is $\omega_{0k}$ if $x=r_{00}$ and $r_{00},r_{0k}\notin S$, but then taking any valid grid and swapping cells $s_{0000}$ and $s_{00k0}$ shows that $S\nvDash x$ and we are done. Thus we may assume that the first two sums in $\psi$ vanish on $S\cup\{x\}$, and therefore the third one vanishes on $S$ but not on $x$, i.e., WLOG
$$\psi=\sum_{ij}z_{ij}\xi_{ij}.$$
That $\psi$ vanishes on $S$ is then equivalent to the following conditions on the matrix $Z=(z_{ij})_{i,j< n}$:
1. $z_{ij}=0$ if $b_{ij}\in S$,
2. $\sum_jz_{ij}=0$ if $r_{i0}\in S$,
3. $\sum_iz_{ij}=0$ if $c_{j0}\in S$.
Let us say that an *alternating path* is a sequence $e=e_p,e_{p+1},\dots,e_q$ of pairs $e_m=(i_m,j_m)$, $0\le i_m,j_m< n$, such that
* $i_m=i_{m+1}$ if $m$ is even, and $j_m=j_{m+1}$ if $m$ is odd,
* the indices $i_p,i_{p+2},\dots$ are pairwise distinct, except that we may have $e_p=e_q$ if $q-p\ge4$ is even,
* likewise for the $j$s.
If $m$ is even, the *incoming line* of $e_m$ is the column $c_{j_m0}$, and its *outgoing line* is the row $r_{i_m0}$. If $m$ is odd, we define it in the opposite way. An *alternating path for $S$* is an alternating path $e$ such that $b_{i_mj_m}\notin S$ for every $m$, and either $e_p=e_q$ and $q-p\ge4$ is even ($e$ is an *alternating cycle*), or the incoming line of $e_p$ and the outgoing line of $e_q$ do not belong to $S$.
Every alternating path $e$ induces a matrix $Z_e$ which has $(-1)^m$ at position $e_m$ for $m=p,\dots,q$, and $0$ elsewhere. It is easy to see that if $e$ is an alternating path for $S$, then $Z_e$ satisfies conditions 1, 2, 3.
**Lemma 2:** The space of matrices $Z$ satisfying 1, 2, 3 is spanned by matrices induced by alternating paths for $S$.
**Proof:**
We may assume that $Z$ has integer entries, and we will proceed by induction on $\|Z\|:=\sum_{ij}|z_{ij}|$. If $Z\ne 0$, pick $e_0=(i_0,j_0)$ such that $z_{i_0j_0}>0$. If the outgoing line of $e_0$ is outside $S$, we put $q=0$, otherwise condition 2 guarantees that $z_{i_0,j_1}< 0$ for some $j_1$, and we put $i_1=i_0$, $e_1=(i_1,j_1)$. If the outgoing line of $e_1$ is outside $S$, we put $q=1$, otherwise we find $i_2$ such that $z_{i_2j_1}>0$ by condition 3, and put $j_2=j_1$. Continuing in this fashion, one of the following things will happen sooner or later:
* The outgoing line of the last point $e_m$ constructed contains another point $e_{m'}$ (and therefore two such points, unless $m'=0$). In this case, we let $p$ be the maximal such $m'$, we put $q=m+1$, $e_q=e_p$ to make a cycle, and we drop the part of the path up to $e_{p-1}$.
* The outgoing line of $e_m$ is outside $S$. We put $q=m$.
In the second case, we repeat the same construction going backwards from $e_0$. Again, either we find a cycle, or the construction stops with an $e_p$ whose incoming line is outside $S$. Either way, we obtain an alternating path for $S$ (condition 1 guarantees that $b_{i_mj_m}\notin S$ for every $m$). Moreover, the nonzero entries of $Z_e$ have the same sign as the corresponding entries of $Z$, thus $\|Z-Z_e\|<\|Z\|$. By the induction hypothesis, $Z-Z_e$, and therefore $Z$, is a linear combination of some $Z_e$s. QED
Now, Lemma 2 implies that we may assume that our $\psi$ comes from a matrix $Z=Z_e$ induced by an alternating path $e=e_p,\dots,e_q$. Assume that $G$ is a valid Sudoku grid that has $1$ in cells $s_{i_mj_m00}$ for $m$ even, and $2$ for $m$ odd. Let $G'$ be the grid obtained from $G$ by exchanging $1$ and $2$ in these positions. Then $G'$ violates the following checks:
* $b_{i_mj_m}$ for each $m$.
* If $e$ is not a cycle, the incoming line of $e_p$, and the outgoing line of $e_q$.
Since $e$ is an alternating path for $S$, none of these is in $S$. On the other hand, $\psi(x)\ne0$ implies that $x$ is among the violated checks, hence $S\nvDash x$.
It remains to show that such a valid grid $G$ exists. We can now forget about $S$, and then it is easy to see that every alternating path can be completed to a cycle, hence we may assume $e$ is a cycle. By applying Sudoku permutations and relabelling the sequence, we may assume $p=0$, $i_m=\lfloor m/2\rfloor$, $j_m=\lceil m/2\rceil$ except that $i_q=j_q=j_{g-1}=0$. We are thus looking for a solution of the following grid:
$$\begin{array}{|ccc|ccc|ccc|ccc|ccc|}
\hline
1&&&2&&&&&&&&&&&&\\\\
\strut&&&&&&&&&&&&&&&\\\\
\strut&&&&&&&&&&&&&&&\\\\
\hline
&&&1&&&2&&&&&&&&&\\\\
&&&&&&&&&&&&&&\cdots&\\\\
&&&&&&&&\ddots&&&&&&&\\\\
\hline
2&&&&&&&&&1&&&&&&\\\\
\strut&&&&&&&&&&&&&&&\\\\
\strut&&&&&&&&&&&&&&&\\\\
\hline
\strut&&&&&&&&&&&&&&&\\\\
\strut&&&&\vdots&&&&&&&&&&&\\\\
\strut&&&&&&&&&&&&&&&\\\\
\hline
\end{array}$$
where the upper part is a $q'\times q'$ subgrid, $q'=q/2$.
If $q'=n$, we can define the grid easily by putting $x_{ijkl}=(k+l,j-i+l)$, where we relabel the numbers $1,\dots,n^2$ by elements of $(\mathbb Z/n\mathbb Z)\times(\mathbb Z/n\mathbb Z)$, identifying $1$ with $(0,0)$ and $2$ with $(0,1)$. In the general case, we define $x_{ijkl}=(k+l+a_{ij}-b_{ij},l+a_{ij})$. It is easy to check that this is a valid Sudoku if the columns of the matrix $A=(a_{ij})$ and the rows of $B=(b_{ij})$ are permutations of $\mathbb Z/n\mathbb Z$. We obtain the wanted pattern if we let $a_{ij}=b_{ij}=i+j\bmod{q'}$ for $i,j< q'$, and extend this in an arbitrary way so that the columns of $A$ and the rows of $B$ are permutations.
This completes the proof that $x\notin\span(S\cup V_0)$ implies $S\nvDash x$. The other properties of $\models$ easily follow, in particular we get a description of maximal incomplete sets of checks by means of alternating paths. To sum it up:
**Theorem:** Let $n\ge2$.
* $S\models x$ if and only if $x\in\span(S\cup V_0)$. In particular, $\models$ is a linear matroid.
* All minimal complete sets of checks have cardinality $3n^2-2n$. (One such set consists of all checks except for one row from each band, and one column from each stack.)
* Closed sets of $\models$ are intersections of maximal closed sets, which are complements of Sudoku permutations of the sets
- $\{b_{00},b_{01},b_{11},b_{12},\dots,b_{mm},b_{m0}\}$ for $0< m< n$
- $\{c_{00},b_{00},b_{01},b_{11},b_{12},\dots,b_{mm},r_{m0}\}$ for $0\le m< n$
- $\{c_{00},b_{00},b_{01},b_{11},b_{12},\dots,b_{m-1,m},c_{m1}\}$ for $0\le m< n$