The minimum number of switches required is $100$.
This puzzle can be turned into a linear algebra problem over ${\Bbb Z}/2{\Bbb Z}$ by letting the vector $v\in({\Bbb Z}/2{\Bbb Z})^{100}$ record which switches are pressed ($1$ if pressed, $0$ if not), the vector $w\in({\Bbb Z}/2{\Bbb Z})^{100}$ record which lights are initially on ($1$ if on, $0$ if off), and the $100$-by-$100$ matrix $M$ over ${\Bbb Z}/2{\Bbb Z}$ record which switches affect which lights. A given $v$ will then solve the puzzle iff $Mv=w$.
As the questioner points out, $v=(1,\dots,1)^T$ is a solution. To show that this is the only solution, it's enough to show that $M$ is invertible.
Suppose that we press all the switches which are either in row $r$ or in column $c$. If a light is not in row $r$ and not in column $c$, its state will be changed twice. If a light is in row $r$ but not in column $c$, its state will be changed $10$ times, and similarly for a light in column $c$ but not in row $r$. The light at $(r,c)$ will have its state changed $19$ times. Therefore, the only light whose state ends up changing is the one at $(r,c)$. Since any desired light can be toggled, this shows that any configuration of lights can be turned off, so $M$ has full range and so must be invertible.
This puzzle is similar to the electronic game Lights Out.
Let $B'=B\cup\{\infty\}$, where $\infty$ isn't a button. If you partition $B'$ into $k>1$ sets, that determines $k-1$ combinations - the partition with $\infty$ isn't counted - and there are $(k-1)!$ ways to order those combinations into a sequence. So $$\sum_{k=2}^{|B|+1} (k-1)!p_k(B')$$ is the count required, where $p_k(X)$ is the number of ways of partitioning a set $X$ into $k$ non-empty sets.
Now $p_n(X)=S(|X|,k)$ is the Stirling number of the second kind. So if $n=|B|$ then $|B'|=n+1$ and we get the numbers:
$$\sum_{k=2}^{n+1} (k-1)!S(n+1,k)$$
Using data from Wikipedia, we have $S(4,2)=7,S(4,3)=6,S(4,4)=1$. That yields, for $|B|=3$, the formulation:
$$1!\cdot 7 + 2!\cdot 6 + 3!\cdot 1 = 25$$
which is the value posted by user paw in comments, and disagrees with your 45.
For $n=1,2,3$ we get $1,5,25$, which is interesting.
For $|B|=4$, though, we get:
$$1!\cdot 15+2!\cdot 25 +3!\cdot 10 + 4!\cdot 1 = 149$$
Wikipedia's formula for $S(n,k)$ might be useful:
$$S(n,k) =\frac{1}{k!}\sum_{j=0}^k (-1)^{k-j}\binom{k}{j}j^n$$
Yielding:
$$\sum_{k=2}^{n+1}\frac{1}{k}\sum_{j=0}^k(-1)^{k-j} \binom{k}{j}j^{n+1}$$
A slightly different counting approach gives:
$$1+2\sum_{k=2}^n k!S(n,k)$$
For example, when $n=4$ this is $$1+2\left( 2!\cdot S(4,2)+3!\cdot S(4,3)+4!\cdot S(4,4)\right)=1+2\cdot 74=149$$
(The counting argument is as follows. We take a partition of $B$ into $k$ parts. We can use either all $k$ parts as combinations, in any order, or, when $k>1$, we can also use only $k-1$ parts. That gives us $2k!$ answers contributed by this partition.)
This formula might be better because Wikipedia says that:
$$a_n=\sum_{k=0}^{n} k!\cdot S(n,k)$$ is something called the $n$th "ordered Bell number," and the count we are looking for, for $n>0$, is thus equal to $2a_n-1$. Ordered Bell numbers (or Fubini numbers) are listed on the OEIS page as sequence number A000670.
Best Answer
If order doesn't matter then all that matters is how many times each switch is toggled. There are four switches; let's say the first one is toggled $a$ times, the second, $b$ times, the third, $c$ times, the fourth, $d$ times. Since the total amount of toggling is given as at most $10000$, we have $$a+b+c+d\le10000$$ So now consider $10004$ stones lying in a line on the ground, and paint $4$ of them blue. This leaves $10000$ unpainted stones in $5$ groups, corresponding to the numbers $a$, $b$, $c$, $d$, and $10000-(a+b+c+d)$. So, the number of solutions of the displayed inequality is $10004$-choose-$4$, which is very roughly $10000^4$. A somewhat less rough estimate would be $(1/24)(10000^4)$.