For any given $T>0$, $y(t,\epsilon)\rightarrow x(t)$ in probability as $\epsilon\rightarrow0$ uniformly for $t\in[0,T]$
Proof: I will sketch the main steps, then fill in the details later.
$(\text{Eq}.(3)-\text{Eq}.(1))/\epsilon$ gives $$dy=-(k_0y+k_1(x_0-1)+\epsilon k_1y)\,dt+(\eta_0y+\eta_1x_0+\epsilon \eta_1y)\,dB \tag5$$ $\text{Eq}.(5)-\text{Eq}.(4)$ gives $$dz = -(k_0z+\epsilon k_1y)\,dt+(\eta_0z+\epsilon \eta_1y)\,dB \tag6$$ where $z=y-x_1$.
Let $\omega$ be an element in the sample space. I am going to prove for any given $T>0$, $$ \mathbf E\big[y(t,\omega,\epsilon)^2\big]<M\quad \text{and}\quad\mathbf E\big[z(t,\omega,\epsilon)^2\big]<\epsilon^2Be^{AT},\quad \forall\epsilon\in(0,\epsilon_0),\ t\in[0,T] $$ for some positive constants $\epsilon_0,\,M,\,A,\,B$. Therefore by the Chebyshev-Markov inequality, we have $z(t,\omega,\epsilon)\rightarrow0$ in probability as $\epsilon\rightarrow0$ uniformly for $t\in[0,T]$.