Number of units in a commutative ring

Your link actually gives a huge hint as to how to do this, as it shows that $a=-a$ in any counterexample, and more generally that $U(R)$ has no subgroups of order $2$.

Assume that $R$ is a communitive ring with unity and $5$ units. Our five units are $1, a, a^{-1}, b, b^{-1}$ for some $a,b$. To see that $a,a^{-1}$ are distinct, notice that, for all $x\in R$,

$$a=a^{-1}\Rightarrow xa=xa^{-1} \Rightarrow xa^{2}=x\Rightarrow a^2=1$$

Thus if $a\neq 1$, then $\{1,a\}\leq U(R)$ has order $2$, which is a contradiction. To see that $a\neq b, a\neq b^{-1}$ just note that if this wasn't to hold, then we just did a bad job of picking $b$, and there are two other units out there by assumption.

Now you need to show that no matter how multiplication is defined between the units, you can always derive a contradiction. Try looking at the possible values of $ab$ and then constructing an element whose square is $1$.