Let $A$ be a unital Banach algebra. Denote by $\mathrm{Inv}(A)$ the invertible elements in $A$, and $\mathrm{Inv}_\ell(A)$ the left-invertible elements. That is, $a \in \mathrm{Inv}_\ell(A)$ if and only there exists $b \in A$ such that $ba=1$.
Question: Is $\mathrm{Inv}(A)$ closed in $\mathrm{Inv}_\ell(A)$?
Note: $\mathrm{Inv}(A)$ is obviously open in $\mathrm{Inv}_\ell(A)$ since it is already open in $A$.
Motivation: If $A = B(X)$, the algebra of bounded operators on some Banach space $X$, then this seems to be true [Edit: Perhaps not! see the below comment of Nate Elredge]. Suppose $a_n \in B(X)$ converge (say in operator norm, but the same argument works if $a_n \to a$ in the strong operator norm) to $a \in B(X)$ where each $a_n$ is invertible and $a$ is left-invertible. Obviously $a$ is injective (it has a left-inverse). Since each $a_n$ is surjective and $a_n \to a$, it is quite easy to see that the range of $a$ must be dense in $X$. On the other hand, the range of $a$ is closed (again, this is because $a$ has a bounded left-inverse), so the range of $a$ is all of $X$. Since $a$ is bijective, it has an inverse in $B(X)$ by the bounded inverse theorem. Or, avoiding the latter technology, it follows easily that any (bounded) left-inverse of $a$ is actually the inverse of $a$.
Your proof can be generalised easily to arbitrary unital Banach algebras.
Let $A$ be a unital Banach algebra and consider the left regular representation of $A$ on itself, that is, the map $\lambda_a x = ax\;(a,x\in A)$. Since $A$ is unital, $\lambda\colon A\to B(A)$ is isometric.
Now, let $(a_n)_{n=1}^\infty$ be a sequence of invertible elements in $A$ which converges to some left-invertible $a\in A$. In particular, $\lambda_{a_n} \to \lambda_a$ as $n\to \infty$ in the norm topology of $B(A)$. By your argument, $\lambda_a$ is invertible in $B(A)$. Let $T\in B(A)$ be such that $T\lambda_a = \lambda_a T = I_A$. This means that for each $\xi \in A$ we have
$$T (a\cdot \xi ) = a \cdot T\xi = \xi.$$
In particular,
$$a\cdot T1 = 1,$$
so $T1$ is a right-inverse for $a$ in $A$. Since $a$ is both left- and right-invertible, it is invertible.
Edit. Mike, everything's ok. The fact you need is Theorem 2.9 in Aliprantis' Introduction to operator theory.