Background
From Wikipedia, if A is an $m\times n$ matrix and B is a $p\times q$ matrix, the the Kronecker product $\mathbf A\otimes \mathbf B$ is the $mn\times nq$ block matrix
$$\mathbf A\otimes \mathbf B=
\begin{pmatrix}
a_{11} \mathbf B & \cdots & a_{1n}\mathbf B \\
\vdots & \ddots & \vdots \\
a_{m1} \mathbf B& \cdots & a_{mn}\mathbf B
\end{pmatrix} $$
Conventionality, the Kronecker sum is defined as follows: if A is an $m\times m$ matrix and B is an $n\times n$ matrix, then
$$\mathbf A\oplus \mathbf B=\mathbf A\otimes \mathbf I_n+\mathbf I_m\otimes \mathbf B$$
which can also be expressed in terms of the matrix exponential
$$\exp(\mathbf A\oplus \mathbf B)=\exp(\mathbf A) \otimes \exp(\mathbf B)$$
What I did
I define the Kronecker sum as follows:
$$\mathbf A\oplus \mathbf B=
\begin{pmatrix}
a_{11} +\mathbf B & \cdots & a_{1n}+\mathbf B \\
\vdots & \ddots & \vdots \\
a_{m1} +\mathbf B& \cdots & a_{mn}+\mathbf B
\end{pmatrix} $$
This worked for my problems where A and B were row and column vectors, respectively.
In order to test the general case I took A as an $m\times n$ matrix and B as an $p\times q$ matrix, where $(m,n,p,q)$ are randomly selected, and I populated these matrices with random complex numbers.
I then calculated $\mathbf A\oplus \mathbf B$ with my formulation and compared it with (the ordinary exponential)
$$\mathbf A\oplus \mathbf B=\ln(e^{\mathbf A} \otimes e^{\mathbf B})$$
In all cases the results were identical to with machine precision. Which bring us to the question: why is the Kronecker sum defined for square matrices?
Computation details
For all calculations $(m,n,p,q)\in[50\ 125]$ in order to run the calculations on a GPU, which was better than six times faster than the CPU.
Best Answer
Okay, I see what you did. Your $\oplus$ and $\otimes$ yield take input tuples of size $a$ and $b$ and produce an $ab$-tuple given by all elementwise sums/products (although you've phrased their definition as though they acted on matrices). With that in mind, it's not surprising that elementwise exponentiation transforms $\oplus$ into $\otimes$.
However, that means that your $\oplus$ is not the Kronecker sum: it does not coincide with the usual definition of Kronecker sum, and it does not transform into the Kronecker product under matrix exponentiation.