Why is stable equivalence necessary in topological K-theory?

It comes down to whether or not you want $K(X)$ to be a group. If all you care about is isomorphism classes of vector bundles, that's a perfectly good monoid. People wouldn't call the monoid $K$-theory because $K$-theory of stuff generally means you've performed some kind of group-completion of a monoid or a category or something.

So there's two ways to respond to your question. The simplest would be "why don't vector bundles have inverses?" which on the most superficial level has the answer not all vector bundles are $0$-dimensional.

Another way to respond would be, what if we performed the formal group-completion of the monoid of isomorphism classes of vector bundles? Well, you get the same thing. General nonsense says you have a map from the group completion to $K(X)$ defined your way, since bundles over reasonable spaces have complementary bundles.

Is this roughly an answer to your question?

edit in response to your comment: Say $X$ is a finite-dimensional CW-complex, and $\epsilon$ is a finite-dimensional vector bundle over $X$. Then there is a classifying map for $\epsilon$, this is a map $X \to G_{\infty,k}$ where $k$ is the dimension of $\epsilon$. $G_{\infty,k}$ is the Grassmannian of $k$-dimensional subspaces of $\mathbb R^\infty$. A basic theorem about $G_{\infty,k}$ is that any map of a finite-dimensional CW-complex into $G_{\infty,k}$ is homotopic to a map into $G_{m,k}$ for $m$ perhaps very large. $G_{m,k}$ is canonically isomorphic to $G_{m,m-k}$ (take orthogonal complements). The bundle classified by the corresponding map $X \to G_{m,m-k}$ is the complementary bundle to $\epsilon$, denote it $\epsilon'$. By design, $\epsilon \oplus \epsilon'$ is a trivial bundle as it's the pull-back of the tangent bundle to $\mathbb R^m$. The fact that maps $X \to G_{\infty,k}$ are homotopic to maps $X \to G_{m,k}$ has many proofs -- one is from the Schubert cell decomposition of $G_{m,k}$ together with cellular approximation.

That's the idea. I believe complementary bundles exist for more than bundles over finite-dimensional spaces but that's the most convenient argument that comes to mind (have to run off to teach a class now).

2nd edit: Let $\mathcal V(X)$ be the group completion of the monoid $\mathcal M(X)$ of isomorphism classes of finite-dimensional vector bundles over $X$. Let $K(X)$ be the stable isomorphism classes of vector bundles over $X$. $K(X)$ is a group for a fairly broad class of spaces, by the arguments above. So by group-completion formal nonsense there is a map $\mathcal V(X) \to K(X)$. The question is whether this map has an inverse or not. Given an element $[\alpha] \in K(X)$, $\alpha \in \mathcal M(X)$ is a vector bundle over $X$ and is well-defined up to stable equivalence. To $\alpha$ there is an associated element of $\mathcal V(X)$ and the question is, is the map

$$K(X) \to \mathcal V(X)$$

well-defined, where you map $[\alpha]_{K(X)} \longmapsto [\alpha]_{\mathcal V(X)}$?

Ah, okay, I see the problem now. This isn't well-defined. But things are off only by a tiny thing. $\mathcal V(X)$ contains a copy of the integers as a direct-summand. This is because $\mathcal M(X) \to \mathbb Z$ given by taking the dimension of a bundle is a homomorphism of monoids (using addition in $\mathbb Z$). There is a map back $\{0,1,2,\cdots \} \to \mathcal M(X)$ given by taking trivial bundles over $X$. So in $\mathcal V(X)$ this gives a splitting $\mathcal V(X) = \mathbb Z \oplus \overline{\mathcal V(X)}$ where $\overline{\mathcal V(X)}$ is the "reduced" group-completion of isomorphism types of vector bundles over $X$. There is a well-defined map $K(X) \to \overline{\mathcal V(X)}$ since you can think of $\overline{\mathcal V(X)}$ as being $\mathcal V(X)$ modulo trivial bundles.

Does that make sense now? I haven't thought about this stuff in much detail for years.