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 $\lambda\rightarrow+\infty$.