It is somewhat surprising that there was no answer in terms of generating function as these are quite straightforward. The combinatorial class of set partitions is given by
$$\def\textsc#1{\dosc#1\csod}
\def\dosc#1#2\csod{{\rm #1{\small #2}}}\textsc{SET}(\textsc{SET}_{\ge 1}(\mathcal{Z}))$$
and hence it has the exponential generating function
$$G(z) = \exp(\exp(z)-1).$$
The combinatorial class of set partitions excluding singletons is
$$\textsc{SET}(\textsc{SET}_{\ge 2}(\mathcal{Z}))$$
and hence it has the exponential generating function
$$H(z) = \exp(\exp(z)-1-z).$$
To answer the first question differentiate $H(z)$ to get
$$H'(z) = H(z) \times (\exp(z)-1).$$
Here is a useful observation which I have included in several of my posts.
When we multiply two exponential generating functions of
the sequences $\{a_n\}$ and $\{b_n\}$ we get that
$$ A(z) B(z) = \sum_{n\ge 0} a_n \frac{z^n}{n!}
\sum_{n\ge 0} b_n \frac{z^n}{n!}
= \sum_{n\ge 0}
\sum_{k=0}^n \frac{1}{k!}\frac{1}{(n-k)!} a_k b_{n-k} z^n\\
= \sum_{n\ge 0}
\sum_{k=0}^n \frac{n!}{k!(n-k)!} a_k b_{n-k} \frac{z^n}{n!}
= \sum_{n\ge 0}
\left(\sum_{k=0}^n {n\choose k} a_k b_{n-k}\right)\frac{z^n}{n!}$$
i.e. the product of the two generating functions is the generating
function of $$\sum_{k=0}^n {n\choose k} a_k b_{n-k}.$$
Therefore when we extract coefficients from the differentiated equation for $H(z)$ we get
$$F(n+1) = \sum_{k=0}^{n-1} {n\choose k} F(k) \times 1 =
\sum_{k=0}^{n-1} {n\choose k} F(k).$$
The upper limit is $n-1$ and not $n$ because $\exp(z)-1$ has no constant term.
This is valid for $n\ge 1.$ The base cases are $F(0)=1$ and $F(1)=0.$
For the second part of the question rewrite the equation for $H(z)$ like this:
$$H(z) = \exp(\exp(z)-1)\exp(-z).$$
Using the convolution of exponential generating functions one more time and recognising $G(z)$ from above we obtain that
$$F(n) = \sum_{k=0}^n {n\choose k} B_k (-1)^{n-k} =
(-1)^n \sum_{k=0}^n {n\choose k} (-1)^k B_k.$$
Here is some Maple code to explore these numbers.
with(combinat, bell);
with(combinat, setpartition);
nsb :=
proc(n)
option remember;
local p, admit, s, res, k;
res := 0;
for p in setpartition([seq(k, k=1..n)]) do
admit := true;
for s in p do
if nops(s) = 1 then
admit := false;
break;
fi;
od;
if admit then res := res+1; fi;
od;
res;
end;
F :=
proc(n)
option remember;
if n=0 then return 1 fi;
if n=1 then return 0 fi;
add(binomial(n-1, k)*F(k), k=0..n-2);
end;
q := n -> (-1)^nadd(binomial(n, k)(-1)^k*bell(k), k=0..n);
