 Solution

Suppose $b$ were a left inverse of an algebra element $x$, so $bx=1$. Then $b{x}_n\to 1$, and $b{x}_n$ is invertible for all sufficiently large $n$ by the Neumann series. For such $n$, $b=(b{x}_n)x_n^{-1}$ is a product of invertibles, hence invertible. The equation $bx=1$ then gives $x=b^{-1}$, a contradiction.

If instead $xb=1$, then $x_{n}b\to 1$, so $x_{n}b$ is eventually invertible. Now $b=x_n^{-1}(x_{n}b)$ is invertible, and again $x=b^{-1}$, a contradiction. This proves that a noninvertible limit of invertibles has neither type of one-sided inverse. No finite-dimensional assumption is used.