EDIT: Strengthened Theorem 2.
The answer to the problem as posed is no, unless P=NP:
Theorem 1. Unless P=NP, there is no LP polytope for Horn-SAT that has only integer extreme points and is optimizable in polynomial time.
On the other hand, the natural polytope $P$ given in the post still suffices to solve Horn-SAT via linear programming, as the solution to the LP $\min\{\sum_v x_v : x\in P\}$ will correspond to a (minimal) satisfying assignment for the given formula $\phi$, as long as $\phi$ is satisfiable.
Theorem 2. For any Horn formula $\phi$, the corresponding polytope $P$ (as defined by the constraints in the post) is feasible iff $\phi$ is satisfiable, and the solution to $\arg\min\{ \sum_v x'_v : x\in P\}$ (if it exists) is a satisfying assignment to $\phi$.
[Caveats: by an "LP polytope for Horn-SAT", I mean a polytope with at least the following properties: given a Horn formula $\phi$, a representation of the corresponding polytope can be constructed in polynomial time, and it has a 0/1 variable $x'_i$ for each Boolean variable $x_i$ in $\phi$, and the feasible integer solutions $x'$ to the LP correspond to the satisfying assignments to $x$ (setting each $x_i$ True if $x'_i=1$ and False otherwise). By "optimizable in polynomial time", I mean that, if we extend this polytope to a full LP by adding any linear objective function, the resulting LP is solvable in polynomial time.]
Proof of Theorem 1. Suppose that there is such a linear program. Then Max Independent Set can be solved in polynomial time as follows. Given an instance $G=(V,E)$ of Max Independent Set:
construct a Horn formula $\phi$ with a variable $x_v$ for every vertex $v\in V$, and a clause $(\neg x_v \vee \neg x_u)$ for each edge $(u, v)\in E$
construct the LP polytope for $\phi$, then obtain a full LP by adding objective function maximize $\sum_v x'_v$
solve the LP using standard techniques to obtain an optimal extreme point $x^*$
return the corresponding vertex set $I=\{v\in V : x^*_v = 1\}$
$I$ is an independent set by our assumption that extreme points of the polytope correspond to satisfying assignments to $\phi$ (and by construction of $\phi$ satisfying assignments for $\phi$ correspond to independent sets).
Given any independent set $I'$, there exists (by definition of $\phi$) a corresponding solution to $\phi$, and therefore a corresponding solution $x'$ to the LP (with $x'_v = 1$ iff $v\in I'$), with objective function $\sum_v x'_v = |I'|$. Hence the size of $I$ is $\sum_v x^*_v \ge \sum_v x'_v = |I'|$. So $I$ has maximum size. $~~~\Box$
Proof of Theorem 2. Recall the standard linear-time algorithm for solving Horn-SAT:
while the formula $\phi$ contains any clause consisting of a single positive literal $x_i$:
$~~~~$set $x_i$ true (remove all clauses containing $x_i$, and remove $\neg x_i$ from all clauses that contain $\neg x_i$)
if the resulting formula contains any empty clause, return "not satisfiable"
otherwise, set all remaining variables to false
In the first step, any satisfying assignment must have $x_i$ true, so forcing $x_i$ true does not change the set of satisfying assignments.
If Step 3 executes, the set of satisfying assignments must be empty, so the (original) formula is not satisfiable.
If Step 4 executes, it satisfies all remaining clauses (as each must contain a negated variable), so generates a satisfying assignment.
To prove the theorem, note that the algorithm and the proof of correctness extend directly to the problem of finding a solution to the polytope $P$ defined by the constraints described in the post. Namely, if there is a "clause" constraint $x_i \ge 1$, then every point $x\in P$ must have $x_i=1$, so forcing $x_i=1$ (and adjusting the remaining constraints accordingly) does not reduce the feasible set. And if there is (in Step 3) an "empty clause", the corresponding constraint will be $0 \ge 1$, so the (original) polytope must be infeasible. And otherwise (in Step 4) setting $x_i=0$ for all remaining variables will satisfy all remaining clause constraints.
(We are using here that when we force $x_i=1$ and simplify, the resulting set of constraints is exactly the set of constraints that corresponds to the Horn formula obtained by setting $x_i$ true in formula $\phi$, and then simplifying as described.)
Finally, by the above reasoning, if the given formula $\phi$ is satisfiable, then the algorithm returns the 0/1 point $x^*$ such that, if $x^*_v = 1$, then $x'_v = 1$ in every feasible point $x'\in P$. Consequently, $x^*$ is the feasible point that (uniquely) minimizes $\sum_v x'_v$. It follows that the solution to the LP $\min\{ \sum_v x'_v : x' \in P\}$ will be $x^*$. $~~~\Box$
