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
Take any ring $S$ without identity. If $R$ is any ring with identity, then $R\times S$ does not have identity. Is this what you seek?