Every finite commutative ring with no zero divisors contains a multiplicative identity?

There is an easy argument which shows that a finite integral domain (commutative unital ring with no zero divisors) is a field. Here I wonder whether this result still stands if the term "unital" is dropped.

In other words, can a finite commutative ring with no zero divisors always contain a multiplicative identity? More generally, if this is true, can we even generalize Wedderburn's little theorem: every finite ring with no zero divisors is a field?

With care, you can do it for finite, nonzero, noncommutative rings with no nonzero zero divisors, the last part meaning that it is left and right cancellative.

Let $a\in R$ be nonzero. Then left multiplication by $a$ on elements of $R$ is injective, and since $R$ is finite $a=ax$ for some $x\in R$. Then it also follows that $aa=axa$ and $a=xa$ by multiplication and cancellation (cancellation being possible in a ring without nonzero zero divisors.)

Then for any other $b\in R$, $bxa=ba$ implies $bx=b$ and $axb=ab$ implies $xb=b$ after cancellations.

At this point we're looking at a finite ring with nonzero identity with no nonzero zero divisors, and Wedderburn's little theorem would carry through to show us it is commutative and a field.

Yes, and this is the usual way to state it (answer to your last question).

As for your first question: let $0\neq a\in R$ be any element. Multiply $a$ by all elements of $R$. When multiplying with two different elements, the two products are different, as there are no zero divisors. So we obtain every element as a product, in particular, $a=ax$ for some $x\in R$.

Let $b\in R$ be arbitrary. Then $bxa=bax=ba$, so again, as there are no zero divisors, we have $bx=b$. Thus $x$ is a unit element.

In fact, with a little more care, it is also possible to get rid of the commutativity condition. Check my calculation, locate the place where I used it, and then you can fix it so that it works for arbitrary finite rings.