To show that $R$ is simple, we need to prove that $R$ has no non-trivial two-sided ideals. We will proceed by contradiction and use the given condition that the subring generated by $I \cup \{1\}$ is $R$ for each nonzero ideal $I$.

1. **Assume $R$ is not simple**:
Suppose $R$ has a non-trivial ideal $I$ such that $0 \subsetneq I \subsetneq R$.

2. Direct sum decomposition:
If $R$ can be decomposed as a direct sum of nonzero rings $A \oplus B$, then $A$ and $B$ must be cyclic by considering $I = B$ and $I = A$ respectively and applying the given condition. This contradicts the non-commutativity of $R$.

3. Finite ring structure:
Since $R$ is finite, it can be decomposed into a direct sum of its $p$-subrings using Bezout's lemma. Given that $R$ is indecomposable, its order must be $p^k$ for some $k$.

4. Jacobson radical:
Let $J(R)$ be the Jacobson radical of $R$. Since $R$ is finite, it is Artinian, and thus $R/J(R)$ is semisimple. If $J(R) = 0$, then $R$ is simple because an Artinian ring with a trivial radical is simple.

5. **Case when $J(R) \neq 0$**:
If $J(R) \neq 0$, then $1$ generates $R/J(R)$, making $R/J(R)$ a cyclic field $\mathbb{F}_p$. Thus, $J(R)$ is the unique maximal ideal of $R$, making $R$ a finite local ring.

6. **Properties of $J(R)$**:
Note that $p \in J(R)$. For any $k$ such that $J(R)^k \neq 0$, $R/(J(R)^k)$ is cyclic by the problem condition. If $p \in J(R)^2$, then $J(R) = J(R)^2$ because both would give $\mathbb{F}_p$ when we quotient $R$ by them. However, for Artinian rings $R$, any right $R$-module $M$ satisfies $M \cdot J(R) = \text{rad}(M)$. Thus, $J(R)^2 = \text{rad}(J(R))$.

7. Contradiction:
Since $J(R)$ is finite and Noetherian, it has a proper maximal submodule, leading to a contradiction. Therefore, $p \in J(R) - J(R)^2$.

8. Ideal structure:
We claim $(p) = J(R)$. By the problem condition, $R/(p)$ is cyclic unless $p = 0$, in which case $J(R) = 0$ and $R$ is the field $\mathbb{F}_p$. Since every element of $R/(p)$ has order $p$, we have $R/(p) = R/J(R) = \mathbb{F}_p$, and $(p) \subseteq J(R)$, so the ideals are equal.

9. Cyclic structure:
Each $R/(p^k)$ is cyclic for all $k$ such that $(p^k) \neq 0$. If $(p^{k+1}) \neq 0$, then $(p^k) \neq \text{rad}(p^k) = (p^{k+1})$. If $q$ is the smallest positive integer for which $p^q = 0$, we can inductively show that for all $n < q$, $R/(p^n) = \mathbb{Z}/(p^n)\mathbb{Z}$.

10. Contradiction in ideal structure:
If $(p^{q-1})$ has $k$ elements, then $R$ has $p^{q-1} \cdot k$ elements, and $1$ has order $p^q$. For any nonzero ideal $U$ satisfying $0 < U < (p^{q-1})$, we have $R/U = \mathbb{Z}/(p^q)\mathbb{Z}$. Distinct ideals $U_1$ and $U_2$ must be disjoint, leading to $U_1 + U_2 = (p^{q-1})$. If two distinct such ideals exist, we have $k^2/p^2 = k$, implying $k = p^2$. This would make $R$ commutative, contradicting the non-commutativity of $R$.

11. Conclusion:
There exists at most one ideal between $(p^{q-1})$ and $0$. The ideals of $R$ form a linear total order under inclusion if $J(R) \neq 0$. If $J(R)$ is nonzero, $R$ is an algebra on a free $\mathbb{Z}/(p^q)\mathbb{Z}$ module with basis $1, b_1, \ldots, b_m$. If the set $b_i$ is nonempty, $(p) = (\mathbb{Z}/p\mathbb{Z})^{1+m}$ where $m > 0$, contradicting that $R/(p) = R/J(R) = \mathbb{F}_p$ is a cyclic field. Thus, $R = \mathbb{Z}/(p^q)\mathbb{Z}$ if $J(R)$ is nonzero, contradicting the non-commutativity of $R$.

12. Final conclusion:
Therefore, $J(R) = 0$, and since $R$ is Artinian with a trivial radical and is indecomposable, $R$ is simple.

$\blacksquare$