There are a few ways to justify it.
First, you could look at the motion of the object as it rotates. In 2D, it turns out that all such motion can be decomposed into motion of the CM, and rotation about the CM. Therefore, rotating around the CM itself is the only way to guarantee the CM doesn't move, and hence is the least energetically costly.
Another way is by direct calculus. Suppose some 1D object is made of point masses at positions $x_i$ with masses $m_i$. Then
$$I(x) = \sum m_i (x_i-x)^2$$
is the moment of inertia about $x$. The minimum is attained when the derivative is zero, so
$$0 = 2 \sum m_i (x_i - x)$$
which implies that
$$x = \frac{\sum m_i x_i}{\sum m_i}.$$
This is the definition of the center of mass.
You don't have to, but it makes the equations easier to deal with because you don't have to account for the moment of acceleration terms. See the 2nd part this this answer about deriving Newton's laws on an abitrary point not the center of mass.
So finally the equations of motion of a rigid body, as described by a frame A not on the center of gravity C is (rather messy) $$ \boxed{ \begin{aligned}
\sum \vec{F} &= m \vec{a}_A - m \vec{c}\times \vec{\alpha} + m \vec{\omega}\times\vec{\omega}\times\vec{c} \\
\sum \vec{M}_A &= I_C \vec{\alpha} + m \vec{c} \times \vec{a}_A - m \vec{c} \times \vec{c} \times \vec{\alpha} +\vec{\omega} \times I_C \vec{\omega} + m \vec{c} \times \left( \vec{\omega} \times \vec{\omega} \times \vec{c} \right)
\end{aligned} } $$
The sum of the forces part equates to mass times acceleration of the center of mass. If the COM is not used, these extra terms appear to account for the change.
To help you the laws of motion can be summarized as follows:
- Linear momentum is defined as mass times the velocity of the center of mass $$\vec{p} = m \vec{v}_{C}$$
- Angular momentum at the center of mass is defined as rotational inertia at the center of mass times angular velocity
$$\vec{L}_C = I_C \vec{\omega}$$
- The net forces acting on a body equal the time derivative of linear momentum
$$ \sum \vec{F} = \frac{{\rm d}}{{\rm d}t} \vec{p} = m \vec{a}_C$$
- The net torques acting on a rigid body about the center of mass equal the time derivative of angular momentum at the center of mass $$\sum \vec{\tau}_C = \frac{{\rm d}}{{\rm d}t} \vec{L}_C = I_C \vec{\alpha} + \vec{\omega} \times I_C \vec{\omega}$$
- To transfer these quantities to a different location A with $\vec{r}=\vec{r}_C -\vec{r}_A$ use the following rules
$$\begin{aligned}
\vec{v}_A & = \vec{v}_C + \vec{r} \times \vec{\omega} \\
\vec{a}_A & = \vec{a}_C + \vec{r} \times \vec{\alpha} + \vec{\omega} \times \left( \vec{r} \times \vec{\omega} \right) \\
\vec{L}_A & = \vec{L}_C + \vec{r} \times \vec{p} \\
\sum \vec{\tau}_A & = \sum \vec{\tau}_C + \vec{r} \times \sum \vec{F} \\
\end{aligned}$$
whereas forces, linear momenta, angular velocity and angular acceleration are shared with the entire rigid body and thus do not change from point to point.
To the above you can add the vector form of the parallel axis theorem with
$$ I_A = I_C - m [\vec{r}\times] [\vec{r}\times]$$
where $[\vec{r}\times]$ is the 3×3 skew symmetric matrix for the cross product operator $\begin{pmatrix}x\\y\\z\end{pmatrix}\times = \begin{vmatrix}0&-z&y\\z&0&-x\\-y&x&0\end{vmatrix}$
This comes out of the momentum transformation from C to A, but it is not the complete picture. To see what happens you have to look at the following 6×6 spatial inertia matrix:
$$\begin{aligned}\vec{p} & =m\vec{v}_{C}=m\left(\vec{v}_{A}-\vec{r}\times\vec{\omega}\right)\\
\vec{L}_{A} & =I_{C}\vec{\omega}+m\,\vec{r}\times\left(\vec{v}_{A}-\vec{r}\times\vec{\omega}\right)\\
\begin{pmatrix}\vec{p}\\
\vec{L}_{A}
\end{pmatrix} & =\begin{vmatrix}m & -m\,\vec{r}\times\\
m\,\vec{r}\times & I_{C}-m\,\vec{r}\times\,\vec{r}\times
\end{vmatrix}\begin{pmatrix}\vec{v}_{A}\\
\vec{\omega}
\end{pmatrix}
\end{aligned}$$
Best Answer
Let the body rotate about the $z$-axis, then by the definition of angular momentum
$$\vec{L}=\vec{\omega} I_z.$$
where $\omega$ is the angular velocity about the $z$-axis. So we could take the parallel axis theorem and multiply it by $\omega$:
$$\vec{\omega}I_{z}=\vec{\omega}I_{cm}+\vec{\omega}ma^2$$
Now ponder the terms in it. If I understand the notation in the König theorem correctly, we have that $\vec{L}_{cm}$ is the angular momentum of the centre of mass about the rotation axis (i.e. as if the mass was concentrated at the COM). This is indeed the last term, so:
$$\vec{L}_{cm}=\vec{\omega}ma^2$$
The term $\vec{\omega}I_{cm}$ can then be defined as $\vec{L}'$, which gives the König relation, as the OP required. A further trivial step would be giving $\vec{L}'$ further physical interpretation (e.g. it is the angular momentum about the COM).