Is split-complex $j=i+2\epsilon$? In matrix representation 
imaginary unit 
$$i=\begin{pmatrix}0 & -1 \\ 1 & 0 \end{pmatrix}$$
dual numbers unit
$$\epsilon=\begin{pmatrix}0 & 1 \\ 0 & 0 \end{pmatrix}$$
split-complex unit
$$j=\begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix}$$
Given this definition, does not it follow that
$$j=i+2\epsilon$$
and as such, one of these systems can be fully expressed through others?

If, ignoring the means by which you reached this conclusion (which was well-addressed in epimorphic's answer), we supposed that
$$j=i+2\varepsilon$$
then it follows that (assuming commutativity)
$$j^2=(i+2\varepsilon)^2=i^2 + 4\varepsilon i+4\varepsilon^2$$
which, replacing each by the definition of their square:
$$1=-1 + 4\varepsilon i$$
which only works if we define $\varepsilon i = \frac{1}2$.  This is a pretty long shot from any "reasonable" definition, since our intuition about $i$ and $\varepsilon$ should certainly not lead us to this point. Moreover, making the definition $\varepsilon i = \frac{1}2$ breaks very important properties of multiplication - for instance, it makes the operation not associative since
$$(\varepsilon^2)i\neq \varepsilon(\varepsilon i)$$
$$0i \neq \varepsilon \frac{1}2$$
$$0\neq \frac{1}2\varepsilon$$
which poses a rather major difficulty for algebra. Moreover, the equation $j=i+2\varepsilon$ is not even particularly special; setting $j=i+\varepsilon$ (i.e. by making $\varepsilon$ correspond to twice the matrix representation you suggest - which is an equally valid matrix representation of the dual numbers) yields that we want $\varepsilon i = 2$ - but this doesn't solve the lack of associativity. In fact, if we want associativity, we conclude that $\varepsilon i$ must not be a linear combination of $1$, $i$, and $\varepsilon$ with real coefficients (since $\varepsilon i$ can't be invertible given that $\varepsilon^2 = 0$ and $\varepsilon i$ can't be a multiple of $\varepsilon$ as that would cause $i\cdot i \cdot \varepsilon$ to break associativity) - which implies that $(i+a\varepsilon)^2=1=j^2$ must have no solution, so we can't reasonably express $j$ in such a system.
The fundamental issue with this is that the expression $i+2\varepsilon$ doesn't even make sense without additional structure. Though we can happily add the terms together in a formal sum (i.e. where we write every number as $a+bi+c\varepsilon$ without allowing any simplification) and this sometimes yields meaningful results, this leaves us with the issue of multiplication. Eventually the term $\varepsilon i$ will come up, and, unless we wish to break important properties of multiplication, we have to consider it as an entirely new thing - and we can prove that, in any extension of $\mathbb R$ which still obeys certain algebraic properties, but contains a new element $\varepsilon$ squaring to $0$ and $i$ squaring to $-1$, there would be only two solutions to $x^2=1$, and they are $1$ and $-1$ - there is no extra root $j$, so $j$ cannot meaningfully play into that system.
