Kameryn Williams has already given a very good answer, but perhaps it is worth saying explicitly that there is not 100% consensus on the exact meaning of completed infinity (or actual infinity).

Having said that, it is common practice to formalize "$\mathbb N$ is not a completed infinity" with the formal system $\mathsf{ZF} - \mathsf{Inf} + \neg\mathsf{Inf}$ (i.e., $\mathsf{ZF}$ with the axiom of infinity replaced by its negation), which is bi-interpretable with first-order Peano arithmetic $\mathsf{PA}$. With this interpretation, the answer to your question about whether it would make any mathematical difference to deny that $\mathbb{N}$ is a completed infinity is yes. Specifically, we would lose theorems that are provable in $\mathsf{ZF}$ but not provable in $\mathsf{PA}$, such as the Paris–Harrington theorem.

There are alternatives. As Kameryn Williams mentioned, Linnebo and Shapiro have argued for a novel interpretation of "$\mathbb{N}$ is only a potential infinity." Under their interpretation, one does lose some theorems, but not quite the same ones. There was some discussion of this point recently on the Foundations of Mathematics mailing list.

As for what it means for $V$ to be a completed whole, there is perhaps even less consensus about what that would mean. As you suggest, one possibility is to introduce a distinction between sets and classes and to assert that $V$ is the class of all sets. You could then point to the conservativity of $\mathsf{NBG}$ over $\mathsf{ZFC}$ as evidence that such an assertion does not prove any new theorems about sets. However, some might argue that the assertion that $V$ is a completed whole amounts to the assertion that $V$ is the set of all sets, and that of course leads to a contradiction in a well-known way. The conclusion is that $V$ is not a completed whole. This is arguably how Cantor thought about the Absolute Infinite.

Kameryn Williams has already given a very good answer, but perhaps it is worth saying explicitly that there is not 100% consensus on the exact meaning of completed infinity (or actual infinity).

Having said that, it is common practice to formalize "$\mathbb N$ is not a completed infinity" with the formal system $\mathsf{ZF} - \mathsf{Inf} + \neg\mathsf{Inf}$ (i.e., $\mathsf{ZF}$ with the axiom of infinity replaced by its negation), which is bi-interpretable with first-order Peano arithmetic $\mathsf{PA}$. With this interpretation, the answer to your question about whether it would make any mathematical difference to deny that $\mathbb{N}$ is a completed infinity is yes. Specifically, we would lose theorems that are provable in $\mathsf{ZF}$ but not provable in $\mathsf{PA}$, such as the Paris–Harrington theorem.

There are alternatives. As Kameryn Williams mentioned, Linnebo and Shapiro have argued for a novel interpretation of "$\mathbb{N}$ is only a potential infinity." Under their interpretation, one does lose some theorems, but not quite the same ones. There was some discussion of this point recently (July 2020) on the Foundations of Mathematics mailing list. EDIT: Further discussion took place in early 2023 following a post by Matthias Eberl, announcing two papers of his on the topic.

As for what it means for $V$ to be a completed whole, there is perhaps even less consensus about what that would mean. As you suggest, one possibility is to introduce a distinction between sets and classes and to assert that $V$ is the class of all sets. You could then point to the conservativity of $\mathsf{NBG}$ over $\mathsf{ZFC}$ as evidence that such an assertion does not prove any new theorems about sets. However, some might argue that the assertion that $V$ is a completed whole amounts to the assertion that $V$ is the set of all sets, and that of course leads to a contradiction in a well-known way. The conclusion is that $V$ is not a completed whole. This is arguably how Cantor thought about the Absolute Infinite.