$\newcommand{\la}{\lambda} $ Let us formalize the question as follows:
Take any complex number $\la$. Let $K(s,t):=s\wedge t-st$ for $s,t$ in $[0,1]$. Find $x\in L^2[0,1]$ such that \begin{equation*} x(t)-\la \int_0^1 K(s,t)x(s)\,ds=t \tag{1} \end{equation*} for almost all $t\in[0,1]$.
Note that the kernel $K$ is the correlationcovariance function of the Brownian bridge, with the eigenfunctions $\sin j\pi\cdot$ and the corresponding eigenvalues $1/(j^2\pi^2)$ for natural $j$; see e.g. Section The Brownian bridge. Also, for $t\in[0,1)$ \begin{equation*} t=\sum_{j=1}^\infty b_j\sin j\pi t,\quad\text{where}\quad b_j:=\frac{2(-1)^{j-1}}{j\pi}. \end{equation*} Therefore, equation (1) can be rewritten as \begin{equation*} x(t)=\sum_{j=1}^\infty c_j\sin j\pi t \tag{2} \end{equation*} with \begin{equation*} c_j-\frac\la{j^2\pi^2}\,c_j=b_j, \end{equation*} so that \begin{equation*} c_j=\frac{b_j}{1-\la/(j^2\pi^2)}; \tag{3} \end{equation*} this unique solution $x$ defined by (2)--(3) exists iff $\la\notin\{j^2\pi^2\colon j=1,2,\dots\}$.