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By definition of the modified Bessel function of the first kind $$ I_k(\lambda)=\sum_{n\geq 0}\frac{1}{n!(n+k)!}\left(\frac{\lambda}{2}\right)^{2n+k} $$ so that we get (the sums transpositions are clearly allowed) $$F(\lambda)=e^{-\lambda-1}\sum_{k\geq 1}\sum_{n\geq 0}\frac{\lambda^{k+n}}{n!(k+n)!}=e^{-\lambda-1}\sum_{n\geq 1}a_n\lambda^n \qquad \mbox{where}\qquad a_n=\frac{1}{n!}\sum_{k=0}^{n-1}\frac{1}{k!}.$$ Thus, deriving under the sign sum

$$ F'(\lambda) = e^{-\lambda-1}\Big(1+\sum_{n\geq 1 }[(n+1)a_{n+1}-a_n)]\lambda^n\Big) = e^{-\lambda-1}\sum_{n\geq 0}\frac{1}{(n!)^2}\lambda^n $$ we obtain the closed form $$ F'(\lambda)=e^{-\lambda-1}I_0(2\sqrt{\lambda}). $$ One finally get $$F'(0)=e^{-1}, \quad F'(1)=e^{-2}I_0(2)=e^{-2}\sum_{n\geq0}\frac{1}{(n!)^2}$$ and, using the asymptotic formula when $\lambda\rightarrow+\infty$ for all $k$ $$ I_k(\lambda)=\frac{e^{\lambda}}{\sqrt{2\pi\lambda}}\Big(1+O(\lambda^{-1})\Big), $$ that $$ F'(\lambda)=\frac{e^{2\sqrt{\lambda}-\lambda-1}}{2\sqrt{\pi\sqrt{\lambda}}}\Big(1+O(\lambda^{-1/2})\Big) $$ when $\lambda\rightarrow+\infty$.

By simple computations : The definition of the modified Bessel function of the first kind yields $$ I_k(\lambda)=\sum_{n\geq 0}\frac{1}{n!(n+k)!}\left(\frac{\lambda}{2}\right)^{2n+k} $$ so that we get (the sums transpositions are clearly allowed) $$F(\lambda)=e^{-\lambda-1}\sum_{k\geq 1}\sum_{n\geq 0}\frac{\lambda^{k+n}}{n!(k+n)!}=e^{-\lambda-1}\sum_{n\geq 1}a_n\lambda^n \qquad \mbox{where}\qquad a_n=\frac{1}{n!}\sum_{k=0}^{n-1}\frac{1}{k!}.$$ Thus, deriving under the sign sum

$$ F'(\lambda) = e^{-\lambda-1}\Big(1+\sum_{n\geq 1 }[(n+1)a_{n+1}-a_n)]\lambda^n\Big) = e^{-\lambda-1}\sum_{n\geq 0}\frac{1}{(n!)^2}\lambda^n $$ we obtain the closed form $$ F'(\lambda)=e^{-\lambda-1}I_0(2\sqrt{\lambda}). $$ One finally get $$F'(0)=e^{-1}, \quad F'(1)=e^{-2}I_0(2)=e^{-2}\sum_{n\geq0}\frac{1}{(n!)^2}$$ and, using the asymptotic formula when $\lambda\rightarrow+\infty$ for all $k$ $$ I_k(\lambda)=\frac{e^{\lambda}}{\sqrt{2\pi\lambda}}\Big(1+O(\lambda^{-1})\Big), $$ that $$ F'(\lambda)=\frac{e^{2\sqrt{\lambda}-\lambda-1}}{2\sqrt{\pi\sqrt{\lambda}}}\Big(1+O(\lambda^{-1/2})\Big) $$ when $\lambda\rightarrow+\infty$.

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, $$F'(0)=e^{-1} \quad\mbox{and}\quad F'(1)=e^{-2}\sum_{n\geq0}\frac{1}{(n!)^2}=e^{-2}I_0(2).$$

By definition of the modified Bessel function of the first kind $$ I_k(\lambda)=\sum_{n\geq 0}\frac{1}{n!(n+k)!}\left(\frac{\lambda}{2}\right)^{2n+k} $$ so that we get (the sums transpositions are clearly allowed) $$F(\lambda)=e^{-\lambda-1}\sum_{k\geq 1}\sum_{n\geq 0}\frac{\lambda^{k+n}}{n!(k+n)!}=e^{-\lambda-1}\sum_{n\geq 1}a_n\lambda^n \qquad \mbox{where}\qquad a_n=\frac{1}{n!}\sum_{k=0}^{n-1}\frac{1}{k!}.$$ Thus, deriving under the sign sum

$$ F'(\lambda) = e^{-\lambda-1}\Big(1+\sum_{n\geq 1 }[(n+1)a_{n+1}-a_n)]\lambda^n\Big) = e^{-\lambda-1}\sum_{n\geq 0}\frac{1}{(n!)^2}\lambda^n $$ we obtain the closed form $$ F'(\lambda)=e^{-\lambda-1}I_0(2\sqrt{\lambda}). $$ One finally get $$F'(0)=e^{-1}, \quad F'(1)=e^{-2}I_0(2)=e^{-2}\sum_{n\geq0}\frac{1}{(n!)^2}$$ and, using the asymptotic formula when $\lambda\rightarrow+\infty$ for all $k$ $$ I_k(\lambda)=\frac{e^{\lambda}}{\sqrt{2\pi\lambda}}\Big(1+O(\lambda^{-1})\Big), $$ that $$ F'(\lambda)=\frac{e^{2\sqrt{\lambda}-\lambda-1}}{2\sqrt{\pi\sqrt{\lambda}}}\Big(1+O(\lambda^{-1/2})\Big) $$ when $\lambda\rightarrow+\infty$.

The definition of $I_k$ yields

$$F(\lambda)=e^{-\lambda-1}\sum_{k\geq 1}\sum_{l\geq 0}\frac{\lambda^{k+l}}{l!(k+l)!}=e^{-\lambda-1}\sum_{n\geq 1}a_n\lambda^n$$ where $$a_n=\frac{1}{n!}\sum_{l=0}^{n-1}\frac{1}{l!}.$$ 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 $F'(0)$ and $F'(1)$.

NB: I'm pretty late I know, I've just discovered this blog.

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, $$F'(0)=e^{-1} \quad\mbox{and}\quad F'(1)=e^{-2}\sum_{n\geq0}\frac{1}{(n!)^2}=e^{-2}I_0(2).$$