First solution. An element $x$ of $R$ such that $x^2 = x$ is called idempotent. We show that $1$ is the only unit in $R$. The proof is based on the remark below:
(*) Let $x$ and $y$ be elements of a ring $R$. If $x \neq 1$ is idempotent, then $xyz \neq 1$ and $zyx \neq 1$ for all $z$ in $R$; in particular, $xy$ and $yx$ are both non-invertible.
Indeed, if $xyz = 1$, then $x = xxyz = x^2yz = xyz = 1$, a contradiction. Hence $xyz \neq 1$; similarly, $zyx \neq 1$.
Back to the problem, fix a non-invertible $x$ in $R \setminus \{0\}$; since $R$ is not a skew-field, there exists at least one such. Since $2x$ is not invertible, it is idempotent, so $2x = 4x^2 = 4x$, that is, $2x = 0$.
We now show that $xy = xyx = yx$ for all $y$ in $R$. To prove the first equality, refer to (*) to infer that $xy - xyx = xy(1-x)$ is non-invertible, hence idempotent, so $xy - xyx = xy(1-x)xy(1-x) = xy(x-x^2)y(1-x) = 0$, since $x^2 = x$. Similarly, $yx - xyx = (1-x)yx$ is non-invertible, hence idempotent, so $yx - xyx = (1-x)yx(1-x)yx = (1-x)y(x-x^2)yx = 0$, since $x^2 = x$.

Second solution. As in the previous solution, we show that $1$ is the only unit of $R$.
We first prove that if $u$ is a unit, and $x \neq 0$ is not, then $u+x$ is not a unit. Let $D$
be the set of all non-invertible elements of $R \setminus \{0\}$; since $R$ is not a skew-field, $D$ is
non-empty. If $x$ is a member of $D$, then so is $-x$, and $x = x^2 = (-x)^2 = -x$, so
$2x = 0$. Then $(1+x)^2 = 1+2x+x^2 = 1+x$, so $1+x$ is not a unit (otherwise,
$1+x=1$, so $x=0$, a contradiction). If $u$ is a unit, and $x$ is a member of $D$, then $ux$
and $1+ux$ are both in $D$, and so is $u+x = u(1+u^{-1}x)$.
Let $x$ be a member of $D$, let $y$ be a member of $R$, and write $x(xy) = x^2y = xy$ and $(yx)x = yx^2 = yx$, to infer that $xy$ and $yx$ are both non-units.
We are now in a position to prove that $1$ is the only unit of $R$. Let $u$ be a unit, and let $x$ be a member of $D$. Then $x+ux$ and $x+ xu$ are both non-units, so $(x+ux)^2 = x+ux$ and $(x + xu)^2 = x + xu$. Expand both squares to write $x^2 + xux + ux^2 + (ux)^2 = x + ux$ and $x^2 + x^2u + xux + (xu)^2 = x + xu$, and infer that $ux = xux = xu$. Finally, since $u + x$ is not a unit, $(u + x)^2 = u + x$, so $u^2 = u$; that is, $u = 1$.