Definition of resolvent set

I'm having trouble understanding some subtlety of definition of resolvent set for given bounded operator A everywhere defined on some Hilbert space. Book I use (and many other sources) give the following: $\lambda \in \mathbb C $ is in resolvent set if $ R_{ \lambda} = ( \lambda \mathbb I - A ) ^ {-1} $ exists, is bounded and range of $\lambda \mathbb I -A$ is dense. Now my reasoning begins. This range is also domain of resolvent. Since it is bounded operator on dense domain it can be extended to whole Hilbert space by continuity. $ ( \lambda \mathbb I - A ) R_{\lambda}$ is equal to identity on dense subset so its extension is identity on whole Hilbert space. Similarly $R_{\lambda} (\lambda \mathbb I - A ) $ is identity on whole Hilbert space by definition of resolvent. But this means that $A - \lambda \mathbb I$ is bijection because it has left and right inverse. Therefore its range is actually whole Hilbert space. But if that is the case, why everyone demands it to be merely a dense subset?

Your reasoning is correct, for a continuous (bounded) everywhere defined operator $A$ on a Banach (in particular on a Hilbert) space, the denseness of the range of $\lambda\mathbb{I} - A$ together with the boundedness of the inverse already implies the surjectivity of $\lambda\mathbb{I} - A$, and an equivalent definition of the resolvent set in this setting is

$$\rho(A) = \{\lambda\in \mathbb{C} : \lambda\mathbb{I} - A \text{ is bijective}\},$$

and this definition is also given in the literature.

The advantage of the definition you cited is that that definition can be used unchanged for the case of unbounded (densely defined) closable operators on Banach (or more specifically Hilbert) spaces. In the case of unbounded operators, the denseness of the range and boundedness of the inverse do not imply surjectivity, so then the two phrasings of the definition would not be equivalent.