Does every vector bundle have a 'tensor inverse'? 
For any vector bundle $E$ over a finite-dimensional CW complex, there is a vector bundle $E'$ such that $E\oplus E'$ is trivial.

For a compact Hausdorff base, this is Proposition 1.4 of Hatcher's Vector Bundles and K Theory, while the general case can be deduced from Tom Goodwillie's comments here. Finite-dimensionality of the base is a necessary hypothesis as the tautological line bundle $\gamma^1 \to \mathbb{RP}^{\infty}$ demonstrates (see Example 3.6 of Hatcher). While these references only deal with real vector bundles, the analogous statements are true for complex vector bundles too (the arguments carry over easily).
My question is whether there is an analogue of the above result when direct sum is replaced by tensor product.

Question: Let $E$ be a vector bundle over a finite-dimensional CW complex. Is there a vector bundle $E'' \neq 0$ such that $E\otimes E''$ is trivial?

One case where such a bundle exists is when $E$ is a line bundle: we can take $E'' = E^*$ as $E\otimes E^* \cong \operatorname{End}(E)$ which is a rank one bundle with a nowhere-zero section, namely $\operatorname{id}_E$.
Stiefel-Whitney classes do not provide any obstructions to the existence of such a bundle. More precisely, for any bundle $E$ over a CW complex of dimension $d$, there is a bundle $E'' \neq 0$ such that $w(E\otimes E'') = 1$. To see this, first choose an integer $j$ such that $2^j > d$. Now note that
\begin{align*}
w(E\otimes\varepsilon^{2^j}) &= w(E^{\oplus 2^j})\\ 
&= w(E)^{2^j}\\ 
&= (1 + w_1(E) + \dots + w_d(E))^{2^j}\\ 
&= 1 + w_1(E)^{2^j} + \dots + w_d(E)^{2^j}\\ 
&= 1
\end{align*}
as $\deg w_i(E)^{2^j} = i2^j > d$ for $i > 0$.
Although one can compute Pontryagin classes of tensor products, see here, I haven't been able to establish the existence of a bundle $E'' \neq 0$ with $p(E\otimes E'') = 1$ (or even just $p_1(E\otimes E'') = 0$). You can use Pontryagin classes to show that $E\otimes\varepsilon^{2^j}$ can be non-trivial, i.e. the above trick to kill Stiefel-Whitney classes won't necessarily give rise to a trivial bundle. For example, if $E = T\mathbb{CP}^2$, the bundle $E\otimes\varepsilon^{2^j}$ is non-trivial for all $j \geq 0$ as $p_1(E^{\oplus 2^j}) = 2^jp_1(E) = 2^jp_1(T\mathbb{CP}^2) \neq 0$.
One thing which makes the direct sum analogue easier to attack is the fact that the existence of a bundle $E'$ with $E\oplus E'$ trivial is equivalent to $E$ being a subbundle of a trivial bundle. I am not aware of a condition on $E$ that is equivalent to the existence of a bundle $E'' \neq 0$ with $E\otimes E''$ trivial. I was hoping that the existence of such a bundle could be interpreted in terms of K theory, but I don't think this is possible as the triviality of $E\otimes E''$ doesn't imply the triviality of $(E\oplus\varepsilon^k)\otimes E''$.
 A: Question: "Let $E$ be a vector bundle over a finite-dimensional CW complex. Is there a vector bundle $E′′≠0$ such that $E⊗E′′$ is trivial?"
Answer: Let $S$ be an affine scheme and let $K_0(S)_{\mathbb{Q}}$ be the Grothendieck group of finite rank locally free sheaves on $S$ (with rational coefficients). Two classes $[E],[F]$ are multiplied using the tensor product:
$$[E]*[F]:=[E\otimes_A F].$$
Assume: For any $[E] \neq 0$ there is an $[F]$ with $[E]*[F]:=[E\otimes_A F]=[A^n]=n \in K_0(S)_{\mathbb{Q}}$. It follows $[E]$ is a unit for the multiplication.
Let $0 \neq x:=\sum_i n_i[E_i]\in K_0(S)$ be any non-zero element. It follows $x=[\oplus_i E_i^{\oplus n_i}]:=[E]$. If such an element exist for $[E]$ it follows $x$ is a unit, hence $K_0(S)_{\mathbb{Q}}$ is a field. There are many examples where the Grothendieck ring is not a field.
Example 1: If $\mathbb{P}^n_k$ is projective space over a field $k$, it follows
$$K_0(\mathbb{P}^n_k)_{\mathbb{Q}} \cong \mathbb{Q}[t]/(t^{n+1})$$
which is not a field - it has nilpotent elements. Similar examples exist in the affine situation.
Example 2: In the topological situation if $M$ is a compact topological space and $K_0^t(M)$ is the Grothendieck group of finite rank complex continuous vector bundles on $M$, there is an isomorphism
$$K_0^t(M)_{\mathbb{Q}} \cong H^{2*}(M,\mathbb{Q}),$$
where $H^{2*}(M,\mathbb{Q})$ is singular cohomology of $M$ with rational coefficients, and this ring is not a field in general - it has nilpotent elements: Given $0\neq x\in H^2(M, \mathbb{Q})$ it follows $x^k\in H^{2k}(M, \mathbb{Q})$ and for large $k$ it follows $H^{2k}(M,\mathbb{Q})=0$, hence $x^k=0$ for large $k$.
Note: Any complex projective variety has the homotopy type of a CW complex.
