I will explain why I believe it is not possible to define the Zariski-open subobjects in terms of the Grothendieck topology alone.
First of all, there is a very important conceptual difference between topologies on sets and Grothendieck topologies: topologies on a set tell you about which sets are open, but Grothendieck topologies tell you about which sieves cover.
For topologies on a set, the axiom that the union of open sets is open means the notion of coverage is inherited from the powerset – there is no freedom to change what it means to cover.
By contrast, the raison d'être of Grothendieck topologies is to change the notion of coverage – even if we restrict our attention to subcanonical topologies, this should be clear from the fact that not every epimorphism in the site becomes an epimorphism of sheaves.
(This is why I prefer the "coverage" terminology.)
The paragraph above should be reason enough to be sceptical about the possibility of recovering any notion of open subobject from a Grothendieck topology in general.
For the category of schemes in particular there are additional difficulties.
Recall that a local ring is a ring $A$ that has a unique maximal ideal.
The topological space $\operatorname{Spec} A$ has this property: there is a point whose only open neighbourhood is the entire space.
Thus, the only Zariski-covering sieve on $\operatorname{Spec} A$ is the maximal sieve!
Nonetheless, provided $A$ is not a field, $\operatorname{Spec} A$ does have non-trivial open subspaces.
So any attempt to identify open subschemes in terms of e.g. minimal generating subsets of covering sieves is doomed to failure.
So much for open subschemes.
What about closed subschemes?
In the category of affine schemes, closed immersions are precisely the regular monomorphisms.
This fails already in the category of schemes, because there are non-separated schemes.
But if you take the category of affine schemes as given there is a general procedure that will construct the Zariski coverage, so it is hard to say that having a Grothendieck topology adds any information here.
I think you are already convinced that the answer to your original question is no, but in the comments you mention the possibility of defining a modality whose fixed points are the open subobjects.
It is not clear to me exactly what you mean but there are obstacles here too.
A unary operation $\Box$ on subobjects that can be represented by an endomorphism of the subobject classifier must be pullback-stable.
In particular, if we assume that $\Box$ preserves the top subobject $\top$, it follows that the operation must be inflationary: indeed, given any monomorphism $f : X \to Y$, we have the following pullback square,
$$\require{AMScd}
\begin{CD}
X @>{\textrm{id}_X}>> X \\
@V{\top_X}VV @VV{f}V \\
X @>>{f}> Y
\end{CD}$$
so we must have $f \le \Box f$ in $\textrm{Sub} (Y)$.
Thus $\Box$ cannot be a non-trivial interior operator.
Edit #2: The counterexample below continues to be valid for schemes. We just take the coproduct $S = \bigsqcup_{i \in I} \text{Spec } \mathbb{Z}$ in the category of schemes instead. The only additional step is to verify that product distributes over infinite coproducts in the category of schemes, since this is no longer true by cartesian closure. Your calculation of the global sections of $\mathcal{O}_S[x_1, \dots x_n]$ is incorrect; if you do this calculation for $S$ as defined above you'll get $\mathbb{Z}[x_1, \dots x_n]^I$ as below, not $\mathcal{O}_S(S)[x_1, \dots x_n] = \mathbb{Z}^I[x_1, \dots x_n]$.
Edit: I was careless. In fact this appears to me to be false. The issue is exactly what you said, that tensor products don't distribute over infinite limits in general. More precisely, write $S$ as a colimit $\text{colim}_i \text{Spec } R_i$ of representables. Then
$$\mathcal{O}(S) \cong \lim_i R_i$$
while, using cartesian closure as below, we have
$$\mathcal{O}(S \times \mathbb{A}^n) \cong \lim_i R_i[x_1, \dots x_n]$$
and we have a natural map
$$(\lim_i R_i)[x_1, \dots x_n] \to \lim_i R_i[x_1, \dots x_n]$$
which is not an isomorphism in general. The issue is that an element of the LHS is a polynomial of fixed degree. To be explicit, suppose $S = \bigsqcup_{i \in I} \text{Spec } \mathbb{Z}$ is an infinite disjoint union of points, so the index set $I$ is infinite. Then we are considering the natural map
$$\mathbb{Z}^I[x_1, \dots x_n] \to \mathbb{Z}[x_1, \dots x_n]^I$$
and the image of this map does not contain any sequence $f_i : I \mapsto \mathbb{Z}[x_1, \dots x_n]$ of polynomials whose degrees get arbitrarily large.
You need to use the fact that presheaf categories are cartesian closed. So if $Y, Z : \text{CRing} \to \text{Set}$ are two presheaves on $\text{Aff}$ then there exists an exponential presheaf $[Y, Z]$ determined by the adjunction
$$\text{Hom}(X \times Y, Z) \cong \text{Hom}(X, [Y, Z]).$$
It follows that $\times$ is a left adjoint and hence preserves colimits in both variables, which lets you reduce to the representable case. What this argument does is actually compute the exponential $[\mathbb{A}^n, \mathbb{A}^1]$ to be the presheaf sending a commutative ring $R$ to the polynomial ring $R[x_1, \dots x_n]$, by reducing the verification of the universal property for all presheaves $X$ to representable presheaves only.
Best Answer
Well, if you are restricting your attention to schemes, there is simply not any better option. One way of saying this is that $V(xy)$ is the smallest closed subscheme that contains both $V(x)$ and $V(y)$. This just amounts to the algebraic fact that any element of $\mathbb{C}[x,y]$ which is divisible by both $x$ and $y$ must be divisible by $xy$. Another way to make this precise is to say that $V(xy)$ is the pushout of $V(x)$ and $V(y)$ over their intersection $V(x,y)$ in the category of schemes. So, if you restrict your attention to the the world of schemes, there just does not exist any better candidate to call the union of the two lines.
Your observation that this is not true in the larger sheaf topos is closely related to the fact that the algebraic fact above is not stable under base-change: if you map $\mathbb{C}[x,y]$ to some other $\mathbb{C}$-algebra, then the images of $x$ and $y$ may no longer have the property that any element divisible by both $x$ and $y$ is divisible by $xy$. In the context of schemes, this is the fact that the operation of taking unions of closed subschemes is not stable under base-change. Explicitly, for instance, if you pull back to the diagonal line $V(x-y)$ in $\mathbb{A}^2$, then $V(x)$ and $V(y)$ both become just the reduced point at the origin, but $V(xy)$ becomes a non-reduced thickened version.