I assume that the initial conditions $a_0,a_1,b_0,b_1$ and that $n\to +\infty$.

Let $A(x):=\sum_{n\geq 0} a_n x^n$ and $B(x):=\sum_{n\geq 0} a_n x^n$. Then the recurrence relations become: $$\begin{cases} A(x) - a_1x - a_0 = \lambda x^2 A(x) + \lambda^* x (A(x)-a_0) + \lambda^* x^2 B'(x), \\ B(x) - b_1x - b_0 = \lambda^* x^2 B(x) + \lambda x (B(x)-b_0) - \lambda x^2 A'(x). \end{cases}$$ That is, we have a system of 2 linear first-order ODE: $$ \begin{bmatrix} A'(x)\\ B'(x)\end{bmatrix} = \begin{bmatrix} 0 & -\lambda^* x^{-2}+x^{-1}+\lambda^{*2}\\ \lambda x^{-2} - x^{-1} - \lambda^2 & 0\end{bmatrix} \cdot \begin{bmatrix} A(x)\\ B(x)\end{bmatrix} + \begin{bmatrix} \lambda^*(b_0x^{-2} + (b_1-\lambda b_0)x^{-1})\\ -\lambda(a_0x^{-2} + (a_1-\lambda^*a_0)x^{-1})\end{bmatrix}, $$ which may be analyzed the standard methods.

I assume that the initial conditions $a_0,a_1,b_0,b_1$ and that $n\to +\infty$.

Let $A(x):=\sum_{n\geq 0} a_n x^n$ and $B(x):=\sum_{n\geq 0} a_n x^n$. Then the recurrence relations become: $$\begin{cases} A(x) - a_1x - a_0 = \lambda x^2 A(x) + \lambda^* x (A(x)-a_0) + \lambda^* x^2 B'(x), \\ B(x) - b_1x - b_0 = \lambda^* x^2 B(x) + \lambda x (B(x)-b_0) - \lambda x^2 A'(x). \end{cases}$$ That is, we have a system of 2 linear first-order ODE: $$ \begin{bmatrix} A'(x)\\ B'(x)\end{bmatrix} = \begin{bmatrix} 0 & -\lambda^* x^{-2}+x^{-1}+\lambda^{*2}\\ \lambda x^{-2} - x^{-1} - \lambda^2 & 0\end{bmatrix} \cdot \begin{bmatrix} A(x)\\ B(x)\end{bmatrix} + \begin{bmatrix} \lambda^*(b_0x^{-2} + (b_1-\lambda b_0)x^{-1})\\ -\lambda(a_0x^{-2} + (a_1-\lambda^*a_0)x^{-1})\end{bmatrix}, $$ which may be analyzed with the standard methods.