Suppose $(S, +, \otimes, \rm 0, [1])$ is an algebraic structure, where $S$ is the set of matrices of any orders, $+$ is usual addition of matrix, $\otimes$ is Kronecker product, $\rm 0$ is zero matrix (dimension not defined) and $[1]$ is $1×1$ matrix with the only entry is $1$. Then $(S, +, \otimes, \rm 0, [1])$ seems to be a semiring because it satisfies all the axioms of a semiring except the absorption law, i.e., $x.0=0~\forall ~x\in S$, where $0$ on R.H.S is of higher dimension than that of L.H.S.
My goal left here is to show that zero matrices on L.H.S and R.H.S are both same or equivalent and if so I might be done.
So, could we show these two zero matrices equivalent or the same?
Your structure fails to be a semiring, because $+$ is not a total operation; e.g. you can't add a $2 \times 7$ to a $4 \times 4$ matrix. Did you mean the direct sum $\oplus$ instead? I.e. the direct sum of two such matrices would be a $6 \times 11$ matrix whose top-left corner contains the first matrix, whose bottom-right corner contains the second matrix, and whose remaining entries are zero.
But regarding the question you're asking, I think the idea you're missing is that the dimensions of a matrix are $m \times n$ where $m$ and $n$ can be any natural number: in particular, they can be zero.
The thing you want $0$ to be, I think, is the matrix of dimension $0 \times 0$. This matrix satisfies $0 \otimes x = 0 = x \otimes 0$ for every matrix $x$.