Why is it true that in a commutative von Neumann regular ring, we have that $ax(1+x)=1+x?$
Definition: We say that a unital ring $R$ is von Neumann whenever for every $a \in R,$ there exist an $x \in R$ such that $a = axa.$
After simplifying the LHS, I get that $ax+axx = ax+xax = ax+x,$ but I'm not getting from this how I can obtain $1+x.$
Well, here we go, let's test that against the smallest nonfield von Neumann regular ring $R=F_2\times F_2$ where $F_2$ is the field of two elements.
Let $x=a=(1,0)$. You get
$ax(1+x)=(1,0)[(1,1)+(1,0)]=(0,0)$
and
$1+x=(0,1)$ which are not equal.
From what you've written it really looks like you've unclearly stated your hypotheses, but in this example $axa=a$ and $xax=x$, so it covers pretty much every interpretation of what you wrote.
So, something is wrong with the statement.
If, for example, you meant that $a$ and $x$ have to be related by $a=axa$, then it would be trivial to show
$ax(1+a)=a(1+x)$
which is as close as I could get to the suggested equality.
And if you meant for $xax=x$, that turns into
$ax(1+x)=ax+x=a(1+x)$ also, of course.