Do the zero divisors form an ideal?

Your definition of an ideal seems nonstandard: for most folks, an ideal must also be a subgroup of the additive group of the ring. So, for counterexample to your conjecture, in the ring $\mathbf Z/6\mathbf Z$ there are zero-divisors $2$ and $3$, the sum of which is a unit.

To the second question: yes! As Alex notes in the comments, $2 \in \mathbf Z$ is an example. Any integral domain which is not a field also provides examples. But it's a good pigeonhole exercise to show that you cannot find such an element in a finite ring.

It is not an ideal, as the next example show. Consider the ring $\mathbb R \times \mathbb R$. The elements $(1,0)$ and $(0,1)$ are zero-divisors but their sum is not. Thus, the set of zero-divisors is not closed under sums.

For your second question, consider the polynomial ring in one variable $\mathbb R[x]$, clearly $x$ is neither a zero-divisor or a unit.