By definition $$ I_k(\lambda)=\sum_{n\geq 0}\frac{1}{n!(n+k)!}\left(\frac{\lambda}{2}\right)^{2n+k} $$ so that $$F(\lambda)=e^{-\lambda-1}\sum_{k\geq 1}\sum_{n\geq 0}\frac{\lambda^{k+n}}{n!(k+n)!}=e^{-\lambda-1}\sum_{n\geq 1}a_n\lambda^n \qquad \mbox{where}\qquad a_n=\frac{1}{n!}\sum_{l=0}^{n-1}\frac{1}{l!}.$$ Thus,
$$F'(\lambda)=e^{-\lambda-1}\Big(1+\sum_{n\geq 1 }[(n+1)a_{n+1}-a_n)]\lambda^n\Big)=e^{-\lambda-1}\sum_{n\geq 0}\frac{1}{(n!)^2}\lambda^n$$ and one get, since all these sums transpositions are clearly allowed, $$F'(0)=e^{-1} \quad\mbox{and}\quad F'(1)=e^{-2}\sum_{n\geq0}\frac{1}{(n!)^2}=e^{-2}I_0(2).$$