Of course, without assumptions on the behavior of the eigenvectors, your desired conclusion will not hold. E.g., for $t:=\lambda$, let $$A(t):=\left( \begin{array}{cc} \cos t+3 & \sin t \\ \sin t & 3-\cos t \\ \end{array} \right),\quad L:=\left( \begin{array}{c} 1 \\ 0 \\ \end{array} \right).$$ The eigenvalues of $A(t)$ are $4$ and $2$ for all $t$, whereas $$AL=\left( \begin{array}{c} \cos t+3 \\ \sin t \\ \end{array} \right),$$ and $\sin t$ cannot be represented as $at+b+c/t+d/t^2+\cdots$ for any constants $a,b,c,d,\dots$.

Of course, without assumptions on the behavior of the eigenvectors, your desired conclusion will not hold. E.g., for $t:=\lambda$, let $$A(t):=\left( \begin{array}{cc} 2+\cos t & \sin t \\ \sin t & 2-\cos t \\ \end{array} \right),\quad L:=\left( \begin{array}{c} 1 \\ 0 \\ \end{array} \right).$$ The eigenvalues of $A(t)$ are $3$ and $1$ for all $t$, whereas $$AL=\left( \begin{array}{c} 2+\cos t \\ \sin t \\ \end{array} \right),$$ and $\sin t$ cannot be represented as $at+b+c/t+d/t^2+\cdots$ for any constants $a,b,c,d,\dots$.