I'm afraid that this is more or less a reformulation of what you asked, but: a bounded operator $T$ on a Banach space $X$ is invertible if and only if it is bounded below (i.e. there is some constant $C>0$ such that $||Tx|| \geq C||x||$ for all $x\in X$) and has dense range. Bounded below implies injective, and it also implies that the range is closed; the range is closed and dense, hence is everything.

So these two different types of spectrum distinguish different ways that the operator can fail to be invertible.

I'm afraid that this is more or less a reformulation of what you asked, but: a bounded operator $T$ on a Banach space $X$ is invertible if and only if it is bounded below (i.e. there is some constant $C>0$ such that $||Tx|| \geq C||x||$ for all $x\in X$) and has dense range. Bounded below implies injective, and it also implies that the range is closed; the range is closed and dense, hence is everything.

So these two different types of spectrum distinguish different ways that the operator can fail to be invertible.

Why is this distinction not made for Banach algebras in general? Well, it doesn't make sense to ask this unless the Banach algebra is presented as a subalgebra of operators on some Banach space. I suppose that you could always consider the left-regular representation of the algebra on itself and then interpret those two conditions in that setting. If $A$ is a Banach algebra, let $a \mapsto L_a$ denote the left-regular representation $L_ab = ab$.

If $L_a$ does not have dense range then $\overline{aA}$ is not all of $A$, which condition I don't think can be reduced to anything simpler.

If $L_a$ is not bounded below, that means that there is a sequence $(x_n)$ in $A$ with $||x_n||=1$ such that $ax_n \to 0$, i.e. $a$ is what is called a (left?) topological zero divisor.