I think there is an elementary proof, but let's check it.
I'm relying on:
There always exist maximal proper homogeneous right ideals (by the usual Zorn's Lemma argument.)
The sum of two homogeneous right ideals is again homogeneous.
Suppose $M$ and $N$ are distinct maximal homogeneous right ideals. Then $M+N=R$, and there exists $m+n=1$ with $m\in M$ and $n\in N$. Because of the grading, the grade zero parts must be such that $m_0+n_0=1$, and because $M$ and $N$ are both proper and homogeneous, neither $m_0$ nor $n_0$ can be units of $R_0$. This implies $R_0$ is not local.
By contrapositive then, we have shown if $R_0$ is local, then $R$ is graded local.
This question is intimately tied to the idea of dimension. Given a commutative ring $R$, its (Krull) dimension is defined as the supremum of the lengths of chains of all prime ideals.
The name dimension for this quantity makes sense because the coordinate ring of an $n$-dimensional affine variety has Krull dimension $n$. Localization has a geometric interpretation, too: a prime ideal $\mathfrak{p}$ corresponds to an irreducible variety $V$. The localization $R_\mathfrak{p}$ consists of all rational functions that are defined at all points of $V$.
If $R$ is a local ring and its unique maximal ideal $\mathfrak{m}$ is the only prime ideal, then $R$ has dimension $0$, which, if $R$ is Noetherian, is equivalent to being Artinian.
The geometric interpretation above indicates how to produce a local ring with any number of prime ideals. Given a positive integer $n$, consider $n$-dimensional affine space $\mathbb{A}^n$ over a field $k$ of characteristic zero. Its coordinate ring is $k[x_1, \ldots, x_n]$, and if we localize at the prime (maximal, actually) ideal $(x_1, \ldots, x_n)$, we obtain the local ring $k[x_1, \ldots, x_n]_{(x_1, \ldots, x_n)}$ which consists of all rational functions on $\mathbb{A}^n$ that are defined at the origin. By the correspondence theorem for prime ideals in a localization, this ring has a chain of prime ideals
$$
(0) \subsetneq (x_1) \subsetneq (x_1, x_2) \subsetneq \ldots \subsetneq (x_1, x_2, \ldots, x_n)
$$
of length $n$.
As for an example where $R$ is not a domain: an affine variety is irreducible iff its coordinate ring is a domain, so we need to consider reducible varieties. Let $A = \frac{k[x,y]}{(xy)}$ which is the coordinate ring of the union of the lines $x = 0$ and $y = 0$ in the plane. Let $\mathfrak{m} = (x,y)$ and let
$$
R = A_\mathfrak{m} = \left(\frac{k[x,y]}{(xy)}\right)_{(x,y)}
$$
which is the local ring at the origin. Then $(x), (y)$, and $(x,y)$ are all prime ideals of $R$.
Best Answer
This does not hold in $\mathbb{F}_2$, since $1 \neq 1+1$. If $k$ is a field besides $\mathbb{F}_2$, pick some $b \in k \setminus \{0,1\}$. Then $(x-b) + b$ expresses $x$ as a sum of two units for all $x \neq b$, and $b = (b-1)+1$ is also a sum of two units.
Next, let $R$ be a commutative local ring with maximal ideal $m$. First, suppose $R/m$ is isomorphic to $\mathbb{F}_2$. Then we cannot write $1$ as a sum of two units of $R$, or else we would be able to write $1$ as a sum of two units in $R/m$. Thus, $R$ does not have the desired property. On the other hand, suppose $R/m$ is not isomorphic to $\mathbb{F}_2$. Then, let $x \in R$ be arbitrary, and write $\overline{x}$ (the image of $x$ in the residue field) as a sum of two units in $R/m$. This means that there exist units $a,b \in R$ such that $x-(a+b) \in m$. Let $y = x-(a+b)$. Then $x = (y + a) + b$ expresses $x$ as a sum of two units.
We conclude that a local ring $R$ with maximal ideal $m$ has the property that each element is a sum of two units if and only if $R/m$ is not isomorphic to $\mathbb{F}_2$.