The specific situation is the following:
Let $n>0$ be a natural number, let $X$ be a finite CW complex of dimension $n$, and let $\xi_0,\xi_1$ be oriented real vector bundles of rank $n$ over $X$ such that $\epsilon\oplus\xi_0 \cong \epsilon\oplus \xi_1$ (where $\epsilon$ is the trivial rank 1 bundle). Here the vector bundles $\xi_i$ are considered as maps $X\to BSO(n)$ and the operation "$\epsilon\oplus\cdot$" is interpreted as compostion with the natural map $BSO(n)\to BSO(n+1)$.
The stable isomorphism asserts that there is a homotopy $H:X\times I\to BSO(n+1)$ between the stabilized bundles, and in oder to determine if $\xi_0\cong \xi_1$ one is lead to the commutative diagram
\begin{array}{ccc}
X\times\{0,1\}&\to &BSO(n)\\
\downarrow & & \downarrow \\
X\times [0,1]&\to &BSO(n+1)
\end{array}
where the upper map restricted to $X\times\{i\}$ is $\xi_i$. The obstructions to lifting $H$ in this diagram find themselves in the groups $H^k(X\times[0,1],X\times\{0,1\};\pi_{k-1}(S^n))$, and so (using the suspension iso) the only potentially non-zero obstruction, say $o(\xi_0,\xi_1)$, lies in $H^n(X;\mathbb{Z})$.
So the question itself is "Can we identify this obstruction with something familiar?" (Maybe restricting to the case of $X$ a manifold or Poincare complex is easier)
In the case where $n$ is even, a guess would be $o(\xi_0,\xi_1)=e(\xi_1)-e(\xi_0)$, the difference of the Euler classes. This is certainly a necessary condition, and at least in the case of $X=S^2$ it is also sufficient.
For $n$ odd this guess is definitely not correct: the euler class of any oriented odd-rank real vector bundle is 2-torsion, so must vanish for any rank $n$ bundle over $S^n$; but the tangent bundle of $S^n$ is stably-trivial and not actually trivial for $n\neq 1,3,7$. In this case it's not as clear what a description of the obstruction should be.
(Remark: if $rank(\xi_i)>dim(X)$, then $\xi_i\cong\xi'_i\oplus\epsilon^k$ where $\xi'_i$ has rank equal to $dim(X)$ (this can be shown with an obstruction argument using the connectivity of Stiefel manifolds). Then by another obstruction argument one can show that in fact $\xi'_0\oplus\epsilon \cong \xi'_1\oplus\epsilon$, and so the case $rank(\xi_i)>dim(X)$ is handled by the case $rank(\xi_i)=dim(X)$.)
Best Answer
This is only a comment, but slightly too long. Let me discuss the case $n=5$, as an example of what the additional information beyond the Euler class could look like.
The classification of oriented real rank 5 bundles over CW-complexes of dimension $\leq 5$ was given in
The main result is that the map $$ [X,BSO(5)]\to H^2(X,\mathbb{Z}/2)\oplus H^4(X,\mathbb{Z}/2)\oplus H^4(X,\mathbb{Z}): $$ $$ E\mapsto (w_2(E),w_4(E),p_1(E)) $$ is injective if and only if both the following conditions are satified:
Condition (A): $H^4(X,\mathbb{Z})$ does not have non-trivial elements of order $4$
Condition (B): $Sq^2 H^3(X,\mathbb{Z}/2)=H^5(X,\mathbb{Z}/2)$.
In particular, this result explains how much isomorphism is determined by characteristic classes. In the case of the $5$-sphere, bundles are stably isomorphic if and only if they have the same characteristic classes. In the case of the $5$-sphere, problems come from Condition (B) - I think that there is a secondary characteristic class there which takes values in $H^5/Sq^2 H^3$ and which can detect the tangent bundle. Maybe this can be found in the paper of Peterson and Stein on secondary characteristic classes, but I do not have access to check right now.
Further information on comparison of isomorphism vs. stable isomorphism in dimensions up to $8$ should be in