Let $d >0$, $a_0 \in \mathbb R^{n \times n}$, $a_1 \in L^2([-d,0];\mathbb R^{n \times n})$ and define $A_\lambda \colon \mathbb R^{n} \to \mathbb R^{n}$ by $$A_\lambda= \lambda I -a_0-\int_{-d}^0 e^{\lambda r} a_1(r)dr $$ for every $\lambda>0$. Here $I$ id the identity.
Is it true that if $\lambda$ is big enough then $A_\lambda$ is invertible?
Of course it is injective, i.e. we have $$|A_\lambda x| \geq |\lambda x| - \left |\left (a_0+ \int_{-d}^0 e^{\lambda r} a_1(r)dr \right)x \right| \geq |\lambda x| -( | a_0|+ d |a_1|_{L^2})x \geq 0 $$ for every $x \neq 0$ if $\lambda \geq | a_0|+ d |a_1|_{L^2}$.
Is it also surjective so that we have the invertibility?
In finite dimensions, yes. By the rank-nullity theorem (a.k.a "dimension formula"), a matrix $A:V\to V$ on a finite dimensional vector space has $\dim V=\dim\ker A+\dim A(V)$. Since $A$ injects, $\ker A=\{0\}$ and $\dim\ker A=0$, so $\dim V=\dim A(V)$ and $A(V)\subseteq V$ consequently implies $A(V)=V$, so it surjects, bijects.
Alternatively, finite dimensional matrices can be thought of as functions $\{1,2,\cdots,n\}\to\{1,2,\cdots,n\}$ by their action on sets of basis vectors in the domain and codomain. If it injects, then this action injects, so it surjects, so the matrix itself surjects.
$A_\lambda$ is invertible for $\lambda\gt|a_0|+d|a_1|_{L^2}$.