I will summarize what we are trying to say with an example. Start with a disk of mass ##M## and radius ##R##. Consider Cartesian coordinate axes with the origin at the center of the disk. Let the ##z## axis be perpendicular to the plane of the disk. The ##x## and ##y## axes are, respectively, along "East" and "North." We know that there are no off-diagonal elements and that ##~I_{zz}=\frac{1}{2}MR^2~## and by the perpendicular axes theorem, ##I_{xx}=I_{yy}=\frac{1}{2}I_{zz}.## Thus, the inertia tensor is $$I_{\text{disk}}=
\frac{1}{2}R^2\begin{pmatrix}
\frac{M}{2} & 0 & 0 \\
0 &\frac{M}{2} & 0 \\
0 & 0 & M \end{pmatrix}.$$ We all agree that this is a "balanced" disk that does not "wobble" when allowed to roll. You say that it is balanced because it has no off-diagonal elements. OK, let's introduce off-diagonal elements by adding mass ##m## on the rim of the disk at 45° "North" of "East", i.e. at point ##\frac{R}{\sqrt{2}}\{1,1,0\}.## The inertia tensor of this mass is
$$I_{\text{mass}}=
\frac{1}{2}R^2\begin{pmatrix}
m & m & 0 \\
m &m & 0 \\
0 & 0 & 2m \end{pmatrix}.$$ The inertia tensor of the composite object is the sum of the two, $$
I_{\text{comp}}=I_{\text{disk}}+I_{\text{mass}}=
\frac{1}{2}R^2\begin{pmatrix}
\frac{M}{2}+m & m & 0 \\
m &\frac{M}{2}+m & 0 \\
0 & 0 & M +2m\end{pmatrix}.$$We all agree that the composite object is unbalanced and will wobble when allowed to roll because of the added mass ##m## on the rim. You say that the off-diagonal elements are an indication that it is so. We say not necessarily and here is why.
Suppose we rotate the coordinate axes about the ##z##-axis by 45° so that the added mass is on the new ##x##-axis. This is equivalent to diagonalizing the matrix. In the new frame the inertia tensor of the added mass is
$$I'_{\text{mass}}=
\frac{1}{2}R^2\begin{pmatrix}
0 & 0 & 0 \\
0 &2m & 0 \\
0 & 0 & 2m \end{pmatrix}$$ whilst the inertia tensor of the disk is the same. Thus, in the new frame the inertia tensor of the composite is
$$
I'_{\text{comp}}=I_{\text{disk}}+I'_{\text{mass}}=
\frac{1}{2}R^2\begin{pmatrix}
\frac{M}{2} & 0 & 0 \\
0 &\frac{M}{2}+2m & 0 \\
0 & 0 & M +2m\end{pmatrix}.$$Here we have an inertia tensor that has zero off-diagonal elements (zero product of inertia) for a disk that we have all agreed is unbalanced because of the added mass on the rim. Clearly, your assertion that "unbalance" is indicated by non-zero product of inertia has no merit.
What you
can say merely by looking at the matrix for ##I'_{\text{comp}}## is that the new ##y##-axis is unstable, meaning that if you start this object spinning in free space with initial angular velocity ##\vec{\omega}=\omega~\mathbf{\hat y}##, the direction of the angular velocity will change with respect to time. That's because ##I_{yy}## has an intermediate value ##I_{xx}<I_{yy}<I_{zz}.## If you've studied
Euler's equations for rigid body dynamics, you might have encountered this.
