= Solution
The set $\mathcal O_K$ consists of the elements of $K$ integral over $\mathbb Z$, meaning roots of monic polynomials in $\mathbb Z[X]$.
First prove the <finite-module criterion for integrality> in a slightly stronger form, for any nonzero finitely generated $\mathbb Z$-submodule $M\subseteq K$ stable under multiplication by $\alpha$. Choose generators $m_1,\ldots,m_r$ and write $\alpha m_i=\sum_j a_{ij}m_j$ with <integer> coefficients. Multiply the vector equation $(\alpha I-A)m=0$ by its <adjugate matrix>. This gives $\det(\alpha I-A)m_i=0$ for every generator. Some generator is nonzero and $K$ is a <field>, so $\det(\alpha I-A)=0$. The <polynomial> $\det(XI-A)$ is monic with <integer> coefficients, proving integrality of $\alpha$. This implication needs no prior closure theorem for $\mathcal O_K$.
Conversely, if $\alpha$ satisfies a <monic polynomial> of degree $d$, then $M=\mathbb Z[\alpha]$ is generated by $1,\alpha,\ldots,\alpha^{d-1}$ and is stable under $\alpha$. Every $\beta\in M$ also maps $M$ into itself because it is an <integer> <polynomial> in $\alpha$; the <finite-module criterion for integrality> just proved makes every such $\beta$ integral. Thus this $M$ really lies in $\mathcal O_K$, proving the stated equivalence without circularity.
For two integral elements $\alpha,\beta$, the module spanned by $\alpha^i\beta^j$ for $i<d_\alpha$, $j<d_\beta$ contains one, is finitely generated, and is stable under both elements by their monic equations. It is therefore stable under $\alpha+\beta$ and $\alpha\beta$. The <finite-module criterion for integrality> proves that both are integral. Negatives follow by replacing a multiplication matrix by its negative, and $0,1$ are integral. Hence $\boxed{\mathcal O_K\text{ is a ring}}$.
The <integral basis of the cubic field of discriminant minus 140> can be obtained as follows. For the specified <field>, Eisenstein at two makes $X^3+2X+2$ irreducible, and $\theta$ is integral because the <polynomial> is monic. The <discriminant> of a monic cubic $X^3+rX+s$ is $-4r^3-27s^2$, so
$$
\boxed{\Delta^2(1,\theta,\theta^2)=-4\cdot2^3-27\cdot2^2=-140.}
$$
Here $\Delta^2$ is the printed notation for the signed <discriminant>, namely the square of a Vandermonde product of the conjugates; that product need not be real.
Suppose $\eta=(a+b\theta+c\theta^2)/2$ is integral with the prescribed binary coefficients. The root traces give $\operatorname{Tr}(1)=3$, $\operatorname{Tr}(\theta)=0$, $\operatorname{Tr}(\theta^2)=-4$. Thus $\operatorname{Tr}(\eta)=(3a-4c)/2$ must be an <integer>, forcing $a=0$. If $b=0$, nonzeroness of the coefficient triple forces $c=1$, giving $\eta=\theta^2/2$. If $b=1$, multiply by $\theta$ and use $\theta^3=-2\theta-2$:
$$
\theta\eta=\theta^2/2-c\theta-c.
$$
Closure of the <ring> then again gives $\theta^2/2\in\mathcal O_K$.
But $N_{K/\mathbb Q}(\theta)=-2$, so the <field norm> of $\theta^{-1}$ is $-1/2$. The <field norm> of an <algebraic integer> is an <integer>, proving $\theta^{-1}\notin\mathcal O_K$. Dividing the defining cubic by $2\theta$ gives $\theta^2/2=-1-\theta^{-1}$. Thus $\theta^2/2$ cannot be integral either.
Finally the <discriminant-index formula for an integral lattice> says $-140=[\mathcal O_K:\mathbb Z[\theta]]^2\operatorname{disc}(K)$. The square of the <integer> index divides $140$, so its only possibilities are one and four, giving index one or two. If the index were two, a nonzero coset would supply an integral element $(a+b\theta+c\theta^2)/2$ after reducing coefficients modulo two, with a nonzero binary triple. The preceding argument would force the impossible integrality of $\theta^2/2$. Hence
$$
\boxed{\mathcal O_K=\mathbb Z[\theta],\qquad\{1,\theta,\theta^2\}\text{ is an integral basis}.}
$$
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