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 <i>completed infinity</i> (or <i>actual infinity</i>).

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 <a href="https://mathoverflow.net/a/555">bi-interpretable with first-order Peano arithmetic</a> $\mathsf{PA}$. 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&ndash;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 <a href="https://cs.nyu.edu/pipermail/fom/2020-July/022261.html">Foundations of Mathematics mailing list</a>.
<b>EDIT:</b> Further discussion took place in early 2023 following a <a href="https://cs.nyu.edu/pipermail/fom/2023-January/023710.html">post by Matthias Eberl</a>, 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 <i>set</i> of all sets, and that of course leads to a contradiction in a well-known way. The conclusion is that $V$ is <i>not</i> a completed whole. This is arguably how <a href="https://en.wikipedia.org/wiki/Absolute_Infinite#Cantor%27s_view">Cantor thought about the Absolute Infinite</a>.