For any non-negative integer $n$ the element $\pm (t \pm \sqrt{t^2 - 1})^n$ has norm $1$, and there are no elements $a(t) + b(t) \sqrt{t^2 - 1}$ of norm $-1$ because $a(1)^2 - b(1)^2 \sqrt{1^2 - 1} \ge 0$.
Suppose $a(t) + b(t) \sqrt{t^2 - 1}$ is a unit which is not one of the above units such that $\deg b$ is minimal. Then it has norm $1$, so $a(t)^2 - b(t)^2 (t^2 - 1) = 1$. It is not hard to see that the units $\pm 1$ are the only units for which $b = 0$, so WLOG $b$ is nonzero. This implies that $\deg a = d+1, \deg b = d$ for some non-negative integer $d$, and moreover the leading terms of $a$ and $b$ must agree up to sign. If the leading terms agree, then
$$( a(t) + b(t) \sqrt{t^2 - 1})(t - \sqrt{t^2 - 1}) = (ta(t) - b(t) (t^2 - 1)) + (tb(t) - a(t)) \sqrt{t^2 - 1}$$
is a unit with the property that the coefficient of $\sqrt{t^2 - 1}$ has degree strictly less than that of $b$ which is not on the above list, which contradicts the assumption of minimality. Similarly, if the leading terms of $a$ and $b$ are opposite in sign, then
$$( a(t) + b(t) \sqrt{t^2 - 1})(t + \sqrt{t^2 - 1}) = (ta(t) + b(t) (t^2 - 1)) + (tb(t) + a(t)) \sqrt{t^2 - 1}$$
is a unit with the property that the coefficient of $\sqrt{t^2 - 1}$ has degree strictly less than that of $b$ which is not on the above list, again contradicting the assumption of minimality. So no such units exist.