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

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.

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.