$M\oplus A \cong A\oplus A$ implies $M\cong A$?
Since you wrote in a comment that you know about tensor products and their quotients, here is a hint. Oh, I should say first that I think this is a great problem (I used it myself when teaching a graduate algebra course) because it is a striking application of multilinear algebra when it's hard to see how to even get started using more direct tools, as far as I know.
Hint: Using exterior powers, there is a Kunneth formula $$ \Lambda^2(M_1 \oplus M_2) \cong \Lambda^2(M_1) \oplus (M_1 \otimes_A M_2) \oplus \Lambda^2(M_2) $$ for any two $A$-modules $M_1$ and $M_2$. (There is a corresponding formula for higher exterior powers, but it's not needed here.) Take the second exterior power of both sides of the isomorphism $M \oplus A \cong A \oplus A$ and use the Kunneth formula. Look closely at what you have. Then take second exterior powers again and look closely at what you have.
The phrase you're looking for is that of a stably free module. Your question is answered in the affirmative in Keith Conrad's (as always) very helpful notes on stably free modules here. In particular, note that it's possible to have $A\oplus M\cong A\oplus A\oplus A$ without having $M\cong A\oplus A$, but this is a minimal counterexample: If $A\oplus M\cong A\oplus A$, then $A\cong M$.
Below here is an answer to a different question than the one being asked, which I'm leaving in only because of a comment that it was helpful.
Nope! Take $A=\mathbb{Z}$ and $M$ the direct sum of countably many copies of $\mathbb{Z}$. Then $M\oplus A\cong M\cong M\oplus M$, but $M\not\cong A$.
On the other hand, for some classes of rings (e.g., $A=\mathbb{Z}$, and I think Dedekind domains) it's true that you can "cancel" finitely-generated modules in the way that you want (that is, if one of the factors is the ring itself).
Let $$ \begin{pmatrix}a&b\\ u&v\end{pmatrix}:A\oplus A\to A\oplus M, $$ with $a,b$ in $A$ and $u,v$ in $M$, be an isomorphism. Let $$ \begin{pmatrix}c&f\\ d&g\end{pmatrix}:A\oplus M\to A\oplus A $$ with $c,d$ in $A$ and $f,g:M\to A$, be its inverse. (In this post all maps are $A$-linear.)
It is tedious but straightforward to check that $$ av-bu:A\to M $$ and $$ cg-df:M\to A $$ are inverse isomorphisms.
Edit. I turned this answer into a community wiki because, as pointed out by user26857, the argument had been described a long time ago by Peter LeFanu Lumsdaine in comments under the accepted answer.