In modules over finite rings $Rv = Rw \iff R^\times v = R^\times w$ 
Let $R$ be a finite ring (with unit, possibly non-commutative), and $M$ a left module over $R$. Let $v,w\in M$. Then
  $$Rv = Rw \iff R^\times v = R^\times w.$$

This follows from Lemma 6.4 in Hyman Bass. K-theory and stable algebra. Publications Mathématiques de l'Institut des Hautes Études Scientifiques 22 (1964), 5–60.
The proof uses the Artin-Wedderburn classification theorem. My question is if there is a simpler proof. (The hard part is the direction $\Rightarrow$, of course.)
As a remark, the statement is not true any more for infinite rings. Counterexamples with $M={}_R R$ can be found on page 466 in Irving Kaplansky. Elementary divisors and modules. Transactions of the American Mathematical Society 66 (1949), 464–491. Counterexamle (b) is over a commutative Noetherian ring $R$.
 A: Let $r,s\in R$ with $rv=w$ and $sw=v$. Since $R$ is finite, there exists $n>0$ such that $f=(rs)^n$ and $e=(sr)^n$ are idempotent.  Note that $ev=v$ and $fw=w$.  Let $r'=fre$ and $s'=esf$.  Notice that $r'v=w$.  Also note that right multiplication by $r'$ gives an $R$-module homomorphism $Rf\to Re$.  In fact, it is surjective (as is easily verified using the definitions of $e,f,r'$).  Similarly right multiplication by $s'$ gives a surjective $R$-module homomorphism $Re\to Rf$.  By finiteness of $R$ we conclude both these homomorphisms are isomorphisms.  It follows that $R(1-f)\cong R(1-e)$ by the Krull-Schmidt theorem (I don't know if this counts as more or less elementary than Wedderburn-Artin).  Such an isomorphism is given by right multiplication by an element $x\in (1-f)R(1-e)$.  
Consider $u=r'+x$.  Then $u$ is a unit since right multiplication by $u$ gives an isomorphism from $R=Rf\oplus R(1-f)$ to $R=Re\oplus R(1-e)$ (as it is the direct sum of the two isos $Rf\to Re$ and $R(1-f)\to R(1-e)$) and in a finite ring an element with a one-sided inverse is invertible.    Also $uv=(r'+x)v=(r'+x)ev=r'ev=r'v=w$ because $x\in (1-f)R(1-e)$ implies $xe=0$.  Thus $w\in R^\times v$.  
