If your group has the property that every element has finite order, but there is no upper bound on the orders of the elements, then it is not the additive abelian group of a ring with identity. The reason is that if there were such a ring structure with an identity $1$, then $1$ would have finite additive order $k$, and then for all $a$ in your group, $k\cdot a=(k\cdot1)a=0a=0$, which forces $a$ to have order at most $k$.
For each prime $p$, the Prüfer $p$-group $\mathbb Z(p^\infty)$ is an example of such a group. The quotient group $\mathbb Q/\mathbb Z$ is another. Direct sums (but not direct products) of infinitely many finite cyclic groups of unbounded order would also be examples.
If $A$ is a commutative unital ring, then there are two equivalent definitions of an unital associative $A$-algebra:
(1) A unital ring $R$ with a unital ring homomorphism $f : A \rightarrow Z(R)$,
(2) An $A$-module $R$ together with an $A$-bilinear product that is associate and unital.
Given (1), then one defines the scalar multplication by $a \cdot r := f(a)r$, and given (2), one defines the ring homomorphism by $a \mapsto a \cdot 1_R$.
Therefore, there seems to be two candidate definitions for a (non-unital) associative algebra:
(1') A ring $R$ with a ring homomorphism $f : A \rightarrow Z(R)$,
(2') An $A$-module $R$ together with an associative $A$-bilinear product.
These are not equivalent, and option (2') is the standard definition.
If by associative algebra we mean (2') (the standard definition), then the answer is
Yes, Yes, Yes, Yes.
If by associative algebra we mean (1') (the non-standard definition), then the answer is
No, Yes, No, Yes.
For the "No"s:
Then need not be unique ring homomorphism $\mathbb{Z} \rightarrow Z(R)$:
Take any unital ring $R$ (commutative or not) and consider it as a ring. Then there is a unique unital ring homomorphism $\mathbb{Z} \rightarrow Z(R)$, but there can be other ring homomorphisms, such as
$$
\mathbb{Z} \xrightarrow{0} Z(R).
$$
In particular this shows that (1') and (2') are not equivalent notions.
Best Answer
The same argument without identities still shows that
$$(a + b)(c + d) = ac + ad + bc + bd = ac + bc + ad + bd$$
hence that $ad + bc = bc + ad$ for any $a, b, c, d \in R$. In other words, commutativity holds for any elements in the image of the multiplication map $R \times R \to R$ (which in the presence of identities is always all of $R$). So the only way to break commutativity is to make the image of the multiplication map small somehow.
An easy way to do this is to take the direct product $R = G \times S$ of a nonabelian group $G$ equipped with the zero multiplication and a nonzero ring $S$ with identity.