Stably free module Let $M$ be any $A$-module. Is it true that $M \oplus A \simeq A^{2} $ implies $ M \simeq A $ ?
 A: Yes when $A$ is commutative. The isomorphism $\Lambda^n(M \oplus N) = \oplus_{p+q=n} \Lambda^p(M) \otimes \Lambda^q(N)$ gives $\Lambda^n(M \oplus A) \cong \Lambda^n(M) \oplus \Lambda^{n-1}(M)$.
If $M \oplus A \cong A^2$, then $0 = \Lambda^3 A^2 \cong \Lambda^3 M \oplus \Lambda^2 M$, hence $\Lambda^2 M=0$.
Therefore, $A \cong \Lambda^2 A^2 \cong \Lambda^2 M  \oplus \Lambda^1 M \cong M$.
More generally, when $M \oplus A$ is free of rank $n$, then $M$ is locally free of rank $n-1$ with $\det(M):=\Lambda^{n-1}(M) \cong A$.
A: Yes, when $A$ is commutative! But the obvious generalization is false.
Some context. A module $M$ over a ring $A$ is stably free if $M \oplus A^n \cong A^m$ for some $n, m$. Any such module is necessarily projective. As it turns out, a stably free module is not necessarily free, but the smallest examples when $A$ is commutative occur when $m \ge 3$; see this expository paper by Keith Conrad for details. 
Here is an unsatisfying proof that when $m = 2$ we in fact have $M \oplus A \cong A^2$ implies $M \cong A$. First, recall that direct sum decompositions of a module are naturally in bijection with idempotents in the endomorphism ring of that module (the kernel is one summand and the image is the other). The endomorphism ring of $A^2$ is $M_2(A)$, so we want to find idempotents in $M_2(A)$. Such an idempotent has the form
$$e = \left[ \begin{array}{cc} x & y \\ z & w \end{array} \right]$$
where
$$x = x^2 + yz$$
$$y = y(x + w)$$
$$z = z(x + w)$$
$$w = yz + w^2.$$
The middle two conditions are satisfied if $x + w = 1$ (and then $y, z$ are arbitrary), which we want anyway; otherwise $y = z = 0$ and we get either zero or the identity. Subtracting the first and last conditions gives
$$x - w = x^2 - w^2 = (x - w)(x + w)$$
so it suffices to satisfy the first condition. 
Our goal now is to prove the following:

If $\text{ker}(e)$ is free, then so is $\text{im}(e)$.

Assume that $\text{ker}(e)$ is freely generated by $(p, q)$. Then
$$xp + yq = zp + wq = 0$$
and moreover any other $p', q'$ which satisfy this property are necessarily a multiple of $p, q$. But the kernel clearly contains $(-y, x)$ and $(-w, z)$, hence we can write
$$-y = rp, x = rq, -w = sp, z = sq$$
for some $r, s$, from which we conclude that
$$e = \left[ \begin{array}{cc} rq & -rp \\ sq & -sp \end{array} \right].$$
But we know that $x + w = rq - sp = 1$, from which it follows that $r$ and $s$ generate the unit ideal, hence $\text{im}(e)$ is freely generated by $(q, -p)$. 
A: Using Leavitt path algebras, you can construct a ring such that $R^3\cong R^2$, but $R^2\ncong R$. Thus, $M=R^2$ furnishes you with an example.
You can read all about these algebras in chapter 3 of this online book
