Let $A$ be a finite commutative ring (not assumed to contain an identity). Suppose that $a \in A$ is not a zero-divisor. Then multiplication by $a$ induces an injection from $A$ to itself, which is necessarily a bijection, since $A$ is finite. Thus multiplication by $a$ is a permutation of the finite set $A$, and hence multiplication by some power of $a$ (which by associativity is the same as some power of multiplication by $a$) is the identity permutation of $A$. That is, some power of $a$ acts as the identity under multiplication, which is to say, it is a (and hence the) multiplicative identity of $A$.
In short, if a finite commutative ring $A$ contains a non-zero divisor, then it necessarily contains an identity, and every non-zero divisor in $A$ is a unit.
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
There are no math authorities. There are just conventions, and as far as I can tell the convention that "ring" means "ring without identity" can only be traced back to people who learned algebra using Hungerford.
The main reason to prefer "ring" to mean "ring with identity" is that I am pretty sure it is the statistically dominant convention, although I don't have the statistics to actually back that up. (Unless this is not what you mean by "reason," in which case I'll guess another possible meaning: for most applications, your rings will have identities.)