The following seems to give a reasonable affirmative answer which avoids computing the coordinate ring directly, and replaces condition (2) with the more natural condition that the subset $\Sigma := X(\overline{k})$ in (1) is stable under the action of the Galois group on $\overline{k}^n$.
Let's be cleaner by working more generally over an arbitrary (not necessarily perfect) field $k$ and with geometrically reduced closed subschemes $X$ in a fixed separated $k$-scheme $Y$ locally of finite type. (Note: now affine schemes are gone; can take $Y$ to be an affine space, but this is irrelevant.) The ${\rm{Gal}}(k_s/k)$-stable set $\Sigma = X(k_s)$ in $Y(k_s)$ recovers $X$ as follows. For a $k$-algebra $A$, $X(A)$ is the ${\rm{Gal}}(k_s/k)$-invariants in $X(A_{k_s})$, so we just need to describe $X(A_{k_s})$ as a ${\rm{Gal}}(k_s/k)$-stable subset of $Y(A_{k_s})$. The description in this latter case will be in terms of $\Sigma$, and the ${\rm{Gal}}(k_s/k)$-stability of $\Sigma$ inside of $Y(k_s)$ will ensure that the description we give for $X(A_{k_s})$ is ${\rm{Gal}}(k_s/k)$-stable inside of $Y(A_{k_s})$. That being noted, we rename $k_s$ as $k$ so that $k$ is separably closed and $\Sigma$ is simply a set of $k$-rational points of $Y$ (so the notation is now marginally cleaner). To detect if a point $y \in Y(A)$ lies in $X(A)$, it is equivalent to do so with the local rings of $A$, so we now take $A$ to be local, say with closed point $s$. Then the subset $X(A)$ in $Y(A)$ consists of those $y \in Y(A)$ such that for any local function $f$ near $y(s)$ which vanishes at all points of $\Sigma$ near $y(s)$, necessarily $f(y) = 0$ in $A$. This formulation makes sense for separated algebraic spaces locally of finite type over $k$ (where "local function" is in the sense of the 'etale topology) and it works there too, even though $Y$ is not generally covered by affine opens.
Of course, one can apply this process to any ${\rm{Gal}}(k_s/k)$-stable subset $\Sigma$ of $Y(k_s)$, and what is the meaning of the resulting functor? It represents the Galois descent $X$ of the Zariski closure in $Y_{k_s}$ of $\Sigma$. In general $X(k_s)$ may be larger than $\Sigma$, but nonetheless $\Sigma$ is Zariski-dense in $X_{k_s}$. This is perfectly interesting in practice, regardless of whether or not $\Sigma$ is equal to $X_{k_s}$, since it is what underlies the construction of derived groups, commutator subgroups, images, orbits, and related things in the theory of linear algebraic groups over a general field. For example, the $k$-group ${\rm{PGL}}_n$ is its own derived group in the sense of algebraic groups, but the commutator subgroup of ${\rm{PGL}}_n(k_s)$ is a proper subgroup whenever $k$ is imperfect and ${\rm{char}}(k)|n$.
To give a nifty application, suppose one begins with an arbitrary closed subscheme $X'$ in $Y$ (such as $X' = Y$!), then forms the ${\rm{Gal}}(k_s/k)$-stable set $X'(k_s)$ (which could well be empty, or somehow really tiny), and then applies the above procedure to get a geometrically reduced closed subscheme $X$ in $X'$. What is it? It is the maximal geometrically reduced closed subscheme of $X'$, and one can check its formation is compatible with products (as well as separable extensions $K/k$, such as completions $k_v/k$ for a global field $k$). If $k$ is perfect then $X = X'_{\rm{red}}$, so this is more interesting when $k$ is imperfect. It is especially interesting in the special case when $X'$ is equipped with a structure of $k$-group scheme. Then $X$ is its maximal smooth closed $k$-subgroup, since geometrically integral $k$-groups locally of finite type are smooth. So what? If one is faced with the task of studying the Tate-Shararevich set for such an $X'$ (e.g., maybe $X'$ is a nasty automorphism scheme of something nice) then all that really intervenes is $X$ since it captures all of the local points, so for some purposes we can replace the possibly bad $X'$ with the smooth $X$. (This trick is used in the proof of finiteness of Tate-Shafarevich sets for arbitrary affine groups of finite type over global function fields.) But beware: if the $k$-group $X'$ is connected (and $k$ is imperfect) then $X$ may be disconnected and have much smaller dimension; see Remark C.4.2 in the book "Pseudo-reductive groups" for an example.