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?

No.
What you've done is specify injective unit-preserving homomorphisms $\phi \colon \mathbb C \hookrightarrow M_2(\mathbb R)$, $\psi \colon Cl_{0,1,0}(\mathbb R) \hookrightarrow M_2(\mathbb R)$, and $\rho \colon Cl_{1,0,0}(\mathbb R) \hookrightarrow M_2(\mathbb R)$ by
$$
\phi(i) = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix},
\quad
\psi(\epsilon) = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix},
\quad
\rho(j) = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}.
$$
Here $\mathbb C \cong Cl_{0,0,1}(\mathbb R)$ is the algebra of complex numbers, $Cl_{0,1,0}(\mathbb R)$ the algebra of dual numbers, $Cl_{1,0,0}(\mathbb R)$ the algebra of split-complex numbers, and $M_2(\mathbb R)$ the algebra of $2\times 2$ real matrices. $Cl_{p,q,r}(\mathbb R)$ refers to the real Clifford algebra with an orthogonal basis that consists of


*

*$1$ and

*$p$, $q$, and $r$ other elements whose squares are respectively $1$, $0$, and $-1$.


With the embeddings/homomorphisms defined this way, we do indeed have that $$\rho(j) = \phi(i) + 2 \psi(\epsilon).$$
The problem, as I mentioned in the comments to another one of your posts, is that the embeddings are not canonical: There are infinitely many ways to embed $Cl_{0,0,1}(\mathbb R)$, $Cl_{0,1,0}(\mathbb R)$, and $Cl_{1,0,0}(\mathbb R)$ into $M_2(\mathbb R)$ while preserving the multiplicative unit, with no obvious reason to prefer one over the others. Thus, we can't just claim that $\phi(i)$ is $i$ and so on. And in general, an alternate set of embeddings $\phi', \psi', \rho'$ results in $\rho'(j) \neq \phi'(i) + 2 \psi'(\epsilon)$. 
Here are some examples. For one, note that the dual numbers are "scale-invariant" in the nilpotent "unit" $\epsilon$: For any nonzero real $t$, we could just as well have called $t\epsilon$ our nilpotent "unit" instead of $\epsilon$, the reason being that $(t\epsilon)^2 = 0$. In parallel, there is a family of unit-preserving embeddings $\psi_t \colon Cl_{0,1,0}(\mathbb R) \hookrightarrow M_2(\mathbb R)$ defined by
$$
\psi_t(\epsilon) = \begin{pmatrix} 0 & t \\ 0 & 0 \end{pmatrix}.
$$
The relationship between $\phi(i)$, $\psi_t(\epsilon)$, and $\rho(j)$ is $\rho(j) = \phi(i) + \frac{2}{t}\psi_t(\epsilon)$.
For another, Wikipedia mentions a Jordan-canonical / diagonal embedding of the split-complex numbers into $M_2(\mathbb R)$, which we denote by $\rho'$, defined by
$$
\rho'(j) = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}.
$$
Observe that $\phi(i)$, $\psi(\epsilon)$, and $\rho'(j)$ are linearly independent.
