Notation $R^{\times}=(\pm 1)\times\mathbb{Z}$ Consider the ring $R=\mathbb{Z}[\sqrt{d}]=\{a+b\sqrt{d}|a,b\in\mathbb{Z}\}$ with a square free $d\in\mathbb{Z}$.
What is meant by the notation $$R^{\times}=(\pm 1)\times\mathbb{Z},$$ where $R^\times$ is the group of units in $R$.
 A: This means that there exists a group isomorphism $R^\times \cong \{\pm 1 \} \times \mathbb{Z}$. In fact it will be given by "multiplication": There exists a unit $\varepsilon \in R$ such that every other unit is uniquely of the form $\pm \varepsilon^{n}$ for $n\in \mathbb{Z}$. Note that this is only true if $d>0$. So for finding such an isomorphism try to prove that the smallest unit $\varepsilon > 1$ in $\mathbb{Z}[\sqrt{d}]$ satisfies the following: Every unit in $R$ may be uniquely written as $\pm \varepsilon^{n}$ for $n\in \mathbb{Z}$. Here is a $\textbf{hint}$: For a unit $u$, look at $\pm u, \pm u^{-1}$, to reduce to the case of showing that a unit $x+y\sqrt{d}$ with $x,y \geq 0$ can be written as $\varepsilon^{n}$ for $n \geq 0$.
This is a special case of $\textit{Dirichlet's unit theorem}$ from Algebraic Number Theory:

For $K$ a number field, we have $\mathcal{O}^\ast \cong \mu(\mathcal{O}) \times \mathbb{Z}^{r_1+r_2-1}$, where $r_1$ denotes the number of real embeddings and $r_2$ denotes the number of pairs of complex embeddings, and $\mu(\mathcal{O})$ is the group of roots of unity in $\mathcal{O}$ for an order $\mathcal{O}$ of $K$).

Looking at the proof of this theorem tells you that the isomorphism is given by multiplication, i.e. if $\varepsilon_i$ denotes the image of $(1,e_i)$, where $e_i=(0,\ldots, 0, 1, 0,\ldots, 0)\in \mathbb{Z}^{r_1+r_2-1}$, then $(u,(n_1,\ldots, n_{r_1+r_2-1}))$ is identified with $u\cdot \varepsilon_1^{n_1}\ldots \varepsilon_{r_1+r_2-1}^{n_{r_1+r_2-1}}$.
In your particular case: $K=\mathbb{Q}(\sqrt{d})$, where $d>0$ is a square free integer and $\mathcal{O}=R$; we have $r_1=2$, $r_2=0$. But if a number field $K$ admits a real embedding, then $\mu(\mathcal{O})=\{\pm 1\}$, simply because any root of unity has to be mapped to a root of unity in $\mathbb{R}$.  
A: As suggested in the comments, the statement should probably be $R^\times \cong \{\pm 1\} \times \mathbb{Z}$.  This is basically a special case of Dirichlet's unit theorem.  Writing $R = \mathbb{Z}[\alpha]$, since $d > 0$, then $R$ has $2$ real embeddings, namely $\alpha \mapsto \pm \sqrt{d}$, where $\sqrt{d} \in \mathbb{R}$ is the usual positive square root.  Then Dirichlet's unit theorem implies that $R^\times$ has rank $1$, hence is generated by a fundamental unit $\epsilon$.  Thus
$$
R^\times = \{\pm \epsilon^m : m \in \mathbb{Z}\} \cong \{\pm 1\} \times \mathbb{Z} \, .
$$
