The answer is no. Let $G$ be a cyclic group of order $n$ not divisible by $2$, let $V$ be an irreducible $2$-dimensional representation of $G$, and consider the associated vector bundle $EG\times_G V\to BG$ (which I'll call $V$ again). Then $V$ has trivial Stiefel-Whitney classes since $H^q(BG;\mathbb{Z}/2)=0$ if $q>0$, but $V$ can have non-vanishing Pontryagin class (we have $H^*(BG;\mathbb{Z})=\mathbb{Z}[[x]]/(nx)$ where $x\in H^2$, and $p(V)=1-a^2x^2$ with $a\in \mathbb{Z}/n$ depending on the original representation.) Since the Pontryagin class satisfies Whitney sum up to $2$-torsion, this gives a counterexample: the virtual bundle $-V$ is does not come from a vector bundle, since $p(-V)=(1-a^2x^2)^{-1}$.
Now you ask, what if we also require that the Pontryagin classes are (finitely) invertible? There's probably a counterexample here too, though I don't have one at hand.
Added later. Here's one. (I hope: I keep needing to fix it.) If $G$ is a finite $p$-group, then the "Borel construction" defines a bijection between of: the set $\mathrm{Rep}_n(G)$ of isomorphism classes of real $n$-dimensional representations of $G$ into the set $\mathrm{Bun}_n(G)$ of isomorphism classes of real $n$-dimensional vector bundles over the classifying space $BG$. In representation theory, there are no additive inverses, so non-trivial bundles over $BG$ which come from representations cannot have inverses. So it's enough to find a non-trivial representation $V$ whose characteristic classes all vanish.
Let $G$ be a cyclic group of order $p^2$, where $p$ is an odd prime, generated by an element $\sigma$. Let $L$ be the $1$-dimensional complex representation given by $\sigma|L=e^{2\pi i/p}$, and write $V$ for the real $2$-dimensional vector bundle underlying the complex line bundle $EG\times_G L\to BG$. It appears that the Pontryagin classes vanish: up to signs, the total Pontryagin class is the total Chern class of $V\otimes \mathbb{C}\approx L\oplus \overline{L}$, and we compute $c(V)=c(L)c(\overline{L}) = (1+px)(1-px) = 1-p^2x^2 = 0$.
(The fact about the bijection $\mathrm{Rep}(G)$ into $\mathrm{Bun}(G)$ is non-trivial; it follows from a theorem of Dwyer and Zabrodsky ("Maps between classifying spaces", LNM 1298). I don't know a more elementary proof, but a condition such as "$G$ is a $p$-group" is probably necessary.)
End addition.
There are obstructions in $K$-theory to "inverting" bundles. There are exterior power operators $\lambda^k:KO(X)\to KO(X)$, such that if $[V]$ is the $K$-class of an actual bundle $V$, we have $\lambda^k([V])=[\Lambda^k V]$, the class of the exterior power bundle. The formal sum $\Lambda_t(x)\in KO(X)[[t]]$ given by
$$
\Lambda_t(x)= 1+\lambda^1(x)t+\lambda^2(x)t^2+\dots
$$
has a Whitney formula ($\Lambda_t(x+y)=\Lambda_t(x)\Lambda_t(y)$), coming from the usual decomposition $\Lambda^nV=\sum \Lambda^iV\otimes \Lambda^{n-i}V$. Furthermore, if $x=[V]$ is the class of an honest $n$-dimensional bundle, we must have $\lambda^i([V])=[\Lambda^iV]=0$ for $i>n$, so that $\Lambda_t(x)$ is polynomial.
In this case, trivial bundles don't have trivial $\Lambda$-class; instead, writing "$n$" for the trivial $n$-plane bundle, we have $\Lambda_t(n)=(1+t)^n$. Thus, an isomorphism $V\oplus W= n$ implies an identity $\Lambda_t([V])\Lambda_t([W])=(1+t)^n$.
So an even stronger form of your question is: if $V$ is a vector bundle (over a nice space) such that $\Lambda_t([V])/(1+t)^n \in KO(X)[[t]]$ is a polynomial, must it follow that there exists a bundle $W$ such that $V\oplus W\approx n$? Again, I don't have a counterexample here.
Let $(E,F,\alpha)$ represent an element of the second group. Wlog $\alpha:E\to F$ is defined globally even though it's only an iso outside $K$. Choose global compactly supported sections $s_1,\dots ,s_n$ of the dual of $E$ such that on $K$ they span the bundle; for large enough $n$ this is possible. Let $E'$ be the trivial rank $n$ bundle and interpret these sections as an injective map $s:E\to E'$. We now have a bundle map $(\alpha,s):E\to F\times E'$ which is injective everywhere. Split off the resulting subbundle of $F\times E'$ and call the other part $F'$. This leads to an isomorphism $E\times F'\cong F\times E'$. Replace $(E,F,\alpha)$ by $(E\times E',F\times E',\alpha\times 1)$. Near infinity, where $\alpha\times 1$ is an iso, the composed iso $E\times E'\to F\times E'\to E\times F'$ looks likes $(1, 0)$ on the $E$ part and therefore looks like $(?,\alpha')$ on the $E'$ part where $\alpha'$ is an iso. Make a little adjustment so that it looks like $(0,\alpha')$ on the $E'$ part. This iso $\alpha':E'\to F'$ near infinity yields $(E',F',\alpha')$ equivalent to $(E,F,\alpha)$, but with the bundles trivial near infinity. (In fact $E'$ is trivial globally.)
Best Answer
The proposition you refer to holds for any space homotopy equivalent to a finite dimensional CW complex. Here are the main points.
The property that any vector bundle $E$ over a space has a complementary bundle $E^\prime$ is preserved by homotopy equivalences.
Any finite dimensional CW complex is homotopy equivalent to a smooth manifold. (A classical result of Whitehead shows that any finite dimensional CW complex is homotopy equivalent to a locally finite finite dimensional CW complex. Then simplicial approximation gives a homotopy equivalent finite dimensional simplicial complex. Embed it into a Euclidean space, take a regular neighborhood, and smooth it.)
The total space of any vector bundle $E$ over a smooth manifold $M$ smoothly embeds into a Euclidean space. The normal bundle bundle of $E$ restricted to $M$ is the desired complementary bunlde $E^\prime$.
Note that any topological manifold is homotopy equivalent to a finite dimensional CW complex.