Is this "subgroup packing" polytope integral?
Andrew(the asker) and I had discussed this over email, and we have shown the conjecture is false. The polytope is not integral for Abelian groups, not even for cyclic groups. 

On the positive side. 

Theorem:
For cyclic groups with order $p^kq$, where $p$ and $q$ are primes and $k\in \mathbb{N}$, the incidence matrix of elements and subgroups is totally unimodular.

This is because the family of subgroups is a union of two laminar families.

Hence, this shows the smallest counterexample for cyclic groups must have order at least $2\times 3\times 5 =30$. This actually explains why no small counterexample was found. 

Andrew ran some computations, and found a counterexample for cyclic group of order $30$. 

Counterexample: $x_0=1/2$, $x_2=\frac{30}{2}-\frac{1}{2}=29/2$, $x_3=\frac{30}{3}-\frac{1}{2}=19/2$, $x_5=\frac{30}{5}-\frac{1}{2}=11/2$, and $0$ everywhere else. 
It's not hard to check the point is feasible. Here I rephrase Andrew's proof that this is actually a vertex. 
There are $30$ tight constraints. The whole group constraint, the three
subgroups generated by $2,3$ and $5$ respectively, and non-negativity constraints. Because we have $30$ variables, $x$ is a vertex.

One might wonder if the polytope for $\mathbb{F}_2^n$ is integral for all $n$. Unfortunately, Andrew also found a non-integral of the polytope for $\mathbb{F}_2^4$.