Let $R$ be a ring (with multiplicative identity $1$) and fix $x\in R$. I want to show that
$1+yx$ is left-invertible for all $y\in R$ if and only if $1+yx$ is invertible for all $y\in R$.
We say that $r\in R$ is left-invertible if there exists $s\in R$ such that $sr=1$. Moreover, we say that $r\in R$ is invertible if there exists $s\in R$ such that $sr=1=rs$.
From the definition, one of the directions is trivial. Namely, if $1+yx$ is invertible for all $y\in R$, then $1+yx$ is left-invertible for all $y\in R$. However, I am having trouble with the converse.
Suppose that $1+yx$ is left-invertible for all $y\in R$. Then $$s(1+yx)=1$$ for some $s\in R$. We can rewrite the equation above as $$s = 1- syx.$$ Since $-sy\in R$, $s$ is left-invertible, and thus $$rs=1$$ for some $r\in R$. Putting everything together, $$r = rs(1+yx) = 1+yx.$$ Hence, $1+yx$ is invertible.