We cannot use Engel's theorem (which holds in arbitrary characteristic). Consider the following counterexample for the Heisenberg Lie algebra in characteristic $p>2$. It is has an irreducible representation of dimension $p$, see here.
The assumption of Engel's theorem is that any element of a Lie algebra acts by a nilpotent operator. However, even for an abelian Lie algebra, this need not be true for a representation.
Here you have and alternative (and I think more understandable and full detailed) proof of the result you want to prove.
$\textbf{Proposition}$
Let $\phi^1:G_1\to \text{GL}(V)$ and $\phi^2:G\to \text{GL}(V)$ two group representations of $G_1$ and $G_2$ with dimensions $d_1$ and $d_2$, respectively, over the same vectorial space $V$. Denoting as $D^1(g^1_i)$ y $D^2(g_j^2)$ the matrices of the representations $\phi^1$ and $\phi^2$, then $D^1(g^1_i) \otimes D^2(g_j^2)$ is a representation with dimension $d_1d_2$ of $G_1\times G_2$.
$\textit{Proof}.$
By the properties of the Kronecker product, it followes that
\begin{equation*}
\begin{split}
[D^1(g_i^1)\otimes D^2(g_j^2)][D^1(g_k^1)\otimes D^2(g_l^2)] & =[D^1(g_i^1)D^1(g_k^1)]\otimes [D^2(g_j^2)D^2(g_l^2)]\\
&=D^1(g_i^1g_k^1)\otimes D^2(g_j^2g_l^2).
\end{split}
\end{equation*}
$\textbf{Theorem}$
With the same notation, if the representations $\phi^1$ and $\phi^2$ are irreducible then $D^1(g^1_i) \otimes D^2(g_j^2)$ of hte group $G_1\times G_2$ is also irreducible. Moreover, all the irreducible representations of $G_1\times G_2$ are the direct product of an irreducible representation of $G_1$ times other of $G_2$.
$\textit{Proof}$
It is known that the irreducibility of $\phi^1$ and $\phi^2$ is equivalent to the condition
\begin{equation*}
\sum_{g\in G_1} \chi^1(g)^*\chi^1(g)=|G_1| \quad \text{y} \quad \sum_{h\in G_2} \chi^2(h)^*\chi^2(h)=|G_2|,
\end{equation*}
where $\chi^1$ and $\chi^2$ are the characters of $\phi^1$ y $\phi^2$, respectively. Then,
\begin{equation*}
\begin{split}
|G_1\times G_2|& =|G_1||G_2|=\left(\sum_{g\in G_1} \chi^1(g)^*\chi^1(g)\right)\left(\sum_{h\in G_2} \chi^2(h)^*\chi^2(h)\right)\\
&=\sum_{g\in G_1}\sum_{h\in G_2} \left(\chi^1(g)\chi^2(h)\right)\left(\chi^1(g)^*\chi^2(h)^*\right).
\end{split}
\end{equation*}
On the other hand, using that the characters of the Kronecker product of the two representations is the product of the characters of the two representations, then
\begin{equation*}
|G_1\times G_2|=\sum_{g'\in G_1\times G_2} \chi(g')\chi(g')^*,
\end{equation*}
where $\chi$ are the characters of the representation on the direct product $G_1\times G_2$. Thus, we infer that the representation $D^1(g^1_i) \otimes D^2(g_j^2)$ of the group $G_1\times G_2$ is also irreducible.
For the second part, let $m_1$ and $m_2$ the number of irreducible representations of $G_1$ and $G_2$, respectively, whose dimensions are denoted as $d_{i,1}$ and $d_{j,2}$. Hence,
\begin{equation*}
|G_1|=\sum_{i=1}^{m_1} d_{i,1}^2 \quad \text{y} \quad |G_2|=\sum_{j=1}^{m_2} d_{j,2}^2.
\end{equation*}
The irreducible representations of $G_1\times G_2$ that are obtained as the Kronecker product of irreducible representations of $G_1$ and $G_2$ will have dimension $d_{i,1}d_{j,2}$. Therefore, the sum of the squares of the dimensions $d_k$ of the irreducible representations $G_1\times G_2$ is
\begin{equation*}
\sum_{k=1}^{m_1m_2}d_k^2= \sum_{i=1}^{m_1} \sum_{j=1}^{m_2} d_{i,1}^2d_{j,2}^2 = |G_1||G_2|=|G_1\times G_2|.
\end{equation*}
We conclude that the Kronecker product of the irreducible representations of $G_1$ and $G_2$ determine all the irreducible representacions of $G_1\times G_2$.
Best Answer
The specific reference: look at Theorems 4.3 and 4.4 on page 653 of the revised third edition (does that make it edition 3.5?) of Lang's "Algebra". These give you precisely the statements you want. The proofs are pretty readable, in my opinion, but if you like I will give more details here.
To apply these theorems to your situation, observe that each of the matrix algebras you are looking at is the direct sum of $n$ left ideals, the $i$th of which consists of matrices which have non-zero entries only in the $i$th column. Each of these ideals is simple, and they are all isomorphic as left modules to $K^n$. This is the definition of simple ring given by Lang: it is a semisimple ring (that is, is a direct sum of simple ideals) with only one isoclass of left ideals appearing in the direct sum decomposition. This also shows that the direct sum of two copies of your matrix algebra is semisimple, and Lang's Thm 4.4 gives the statement about its representations that you need.