While I'm sure you know this, it's probably worth mentioning that most of the time, the real root of a given cubic polynomial with one real root won't be a unit at all, let alone fundamental. So for the rest of the discussion let's assume that $P\,$ has constant term $\pm 1$.
There are at least a few results in this direction, all stemming from the fact that by the cubic formula, computations in cubic fields (e.g., roots, discriminants, etc.) can be made very explicit. An excellent summary of these is Found in Frolich + Taylor's chapter on cubic and sextic fields in their Algebraic Number Theory, in which they explicitly address cubic number fields with precisely one real embedding.
I'll just summarize one super-handy lemma due to Artin and an application due to Ishida: Let $K$ be a cubic field with precisely one real embedding, and of discriminant $\Delta$.
If $u>1$ is a unit of $K$ such that $4u^{3/2}+24<|\Delta|$, then $u$ is a fundamental unit.
Back to polynomials, which was the original question: Here's a nice class of polynomials where you can check, very explicitly via the cubic formula, that the one real root of the polynomial satisfies the above bound:
Theorem (Ishida, p.202 in Frolich-Taylor): Suppose $\ell\geq 2$ has the property that $4\ell^3+27$ is square-free, and let $v$ denote the unique real root of $X^3+\ell X-1$. Then $v^{-1}$ is a fundamental unit of $\mathbb{Q}(v)$.
The calculations get somewehat trickier for arbitrary cubic polynomials, but it seems plausible that you could find a similar result for the natural generalization of your class of "nice" polynomials.
Let me just point you to the Wikipedia article on Fundamental units. I think it explains most of your questions so there's no point in repeating what's already in there.
Also, this mathworld article has more information and some examples.
Best Answer
This is true more generally if $\zeta$ is a primitive $n$-th root of unity, where $n$ is odd (not necessarily prime).
Let $u \in \mathbb Z[\zeta]$. Let $\overline{u}$ denote the complex conjugate of $u$ (remark that this is independent of any choice of embedding of $\mathbb Q(\zeta)$ in $\mathbb C$, because $\mathbb Q(\zeta)$ is a CM field - complex conjugation always acts via $\zeta \mapsto \zeta^{-1}$).
Let $s= u/\overline{u}$.
Remark that under any embedding of $\mathbb Q(\zeta)$ in $\mathbb C$, $s$ is mapped to a complex number of absolute value $1$.
It follows that $s$ is a root of unity contained in $\mathbb Q(\zeta)$, hence $s=\zeta^a$ for some $a \in \mathbb Z/n\mathbb Z$.
Let $b \in \mathbb Z/n\mathbb Z$ be such that $2b=-a$, which exists because $n$ is odd. Let $v = \zeta^b u$. Then
$$\overline{v} = \zeta^{-b} \overline{u} = \zeta^{-b} u/s = \zeta^{-b - a} u = \zeta^b u = v.$$
Hence $v$ is totally real, hence $v \in \mathbb Z[\zeta + \zeta^{-1}]^\times$. Since $u = \zeta^{-b}v$, it follows that $\mathbb Z[\zeta]^\times = \left<\zeta\right> \mathbb Z[\zeta + \zeta^{-1}]^\times$.
Remark: This calculation is very Galois-cohomological in flavor. I'd be curious to see a proof of this result using only Galois cohomology!