Isn't it sufficient to give the definition of a group of units
The units of a ring $R$ form a group under multiplication, the group of units of $R$.
to show that $R^{\times}$ form a group under multiplication ?
If not, is it okay to present the neutral element, the inverse element and that the multiplication leads to closure by writing:
Since $(R, +, \cdot)$ is a commutative Ring with $1$, there exists an element $1$ in $R$ with $a \cdot 1 = a $ for all $a \in R^{\times}$. There also exists for each $a \in R^{\times}$ an inverse element $a^{-1}$ so that $a \cdot a^{-1} = 1$. Let $a, b \in R^{\times}$ then $(a \cdot b)(a \cdot b)^{-1} = a \cdot b \cdot a^{-1} \cdot b^{-1} = a \cdot a^{-1} \cdot b \cdot b^{-1} = 1 \cdot 1 = 1 \Rightarrow a\cdot b \in R^{\times} \quad\forall a,b \in R^{\times} $
The statement you've cited is not a definition. It's a theorem followed by a definition. You are asked to prove the theorem contained in that statement.
For instance, here's my definition of the group of primes:
It is rather apparent that the first sentence is false, and therefore the subsequent definition is moot - or, at the very least, it can't hold as it is supposed to.
For the task at hand, you need to prove that:
the restriction of the multiplication to $R^\times \times R^\times$ has image contained in $R^\times$ (important)
that $R^\times$ contains an element $u$ such that $u x=x u=x$ for all $x\in R^\times$ (what might it be?)
the multiplication is associative (easy)
for all $x\in R^\times$, there is $y\in R^\times$ such that $yx=xy=u$ (what might it be?).