Formally write $$u(x,\lambda) = \sum_{j=0}^\infty u_j(x) (\lambda - 1)^j$$ The coefficients of each power of $\lambda-1$ in the differential equation give you a sequence of d.e.'s for the $u_j$'s: $u_0'' + u_0' - i V u_0$ (so that $u(x,1) = u_0(x)$ is a solution of your d.e. for $\lambda = 1$), and $u_j'' + u_j' - i V u_j = i V u_{j-1}$ for $j \ge 1$. Thus $u_j$ is a solution of the inhomogeneous linear equation corresponding to your homogeneous equation, with forcing term depending on $u_{j-1}$.

Of course, to get a particular solution you need to specify initial or boundary conditions, and you have not done so. Convergence or divergence of the series may depend on those conditions.

Presumably $V(x)$ is a continuous function of $x$.

Formally write $$u(x,\lambda) = \sum_{j=0}^\infty u_j(x) (\lambda - 1)^j$$ The coefficients of each power of $\lambda-1$ in the differential equation give you a sequence of d.e.'s for the $u_j$'s: $u_0'' + u_0' - i V u_0$ (so that $u(x,1) = u_0(x)$ is a solution of your d.e. for $\lambda = 1$), and $u_j'' + u_j' - i V u_j = i V u_{j-1}$ for $j \ge 1$. Thus $u_j$ is a solution of the inhomogeneous linear equation corresponding to your homogeneous equation, with forcing term depending on $u_{j-1}$.

Of course, to get a particular solution you need to specify initial or boundary conditions, and you have not done so. Convergence or divergence of the series may depend on those conditions.
If you specify initial conditions that are fixed or entire as functions of $\lambda$, then $u(x,\lambda)$ should be an entire function of $\lambda$, and the power series will always converge.

On the other hand, a solution satisfying specified boundary conditions might fail to exist for certain complex $\lambda$, and presumably this will be reflected in lack of convergence of the power series.