Lighting up all elements of a poset by toggling upsets
Co-worker here. We haven't solved it yet, but here are a few remarks (in case it gives anyone an idea, because we are stuck).

The main thing we have for now is a partial result on so-called crown-free lattices. To show it, for two elements $x,y$ of the poset $P$, I say that $x$ is covered by $y$ if $x\geq y$, $x \neq y$, and there are no elements in between (i.e., $y$ is just above $x$ in the Hasse diagram of $P$). Let me now show claim (*): we can then assume without loss of generality that $P$ is such that:


there are no elements $x$ of $P$ that cover exactly one element; and
for every minimal element $x$ of $P$, $x$ is covered by at least two elements.


I will show this by taking a poset $P$ such that it has an $x$ satisfying condition 1. or 2., and then considering the poset $P'$ where we have removed $x$, and show that if $P'$ is winnable then so is $P$. Assume first that $x$ satisfies 1., and let $\mathbf{s}$ be a winning sequence for $P'$ (a sequence of moves that goes from $\emptyset$ to $P'$). Then it is easy to see that $\mathbf{s}$ is also a winning sequence for $P$; indeed the state of $x$ (on or off) will always follow the state of its unique child, and we never need to click on $x$ (this can also be explained by checking that $\mu_P(x) = 0$), so all moves are valid. Assume now that $x$ satisfies 2., and again let $\mathbf{s}$ be a winning sequence for $P'$. Let $p$ be the unique parent of $x$ in $P$, and let $a = \mu_P(p)$. It is easy to show that $\mu_{P'}(p) = a+1$ and that $\mu_{P'}(y) = \mu_P(y)$ for all $y \in P'$, $y \neq p$. I then consider two cases:


$a&lt;0$. We can then obtain a winning sequence for $P$ as follows: first, play an on-move on $x$, then play an off-move on $p$ (this is legal since $a&lt;0$), and then play the sequence $\mathbf{s}$. Observe that we have $a+1 \leq 0$, so in the whole sequence we will only play off-moves on $p$, and it is clear that all the moves are valid and that we end up with everything being on, so that this is indeed a winning sequence for $P$.
$a \geq 0$. In this case $a+1&gt;0$, which mean that in $\mathbf{s}$ we will only play on-moves on $p$, and we will play at least one. We then obtain a winning sequence for $P$ as follows: play the sequence $\mathbf{s}$, but the very first time we play an on-move on $p$, play an on-move on $x$ instead. Again, it is easy to see that this is indeed a winning sequence for $P$.


This concludes the proof of claim (*).

If we look at the first example in the question, we can deal with it using claim (*): we can eliminate $E$ because it satisfies condition 1., then we can eliminate $A$, $B$ and $C$ (because they satisfy condition 2.), and we end up with only one node ($D$), and this is clearly a poset that is winnable. However for the second example in the question, we can eliminate $G$ (it satisfies condition 2.), but then there is nothing to eliminate, so here he'd need something else to show that $P'$ is winnable.

Claim (*) implies in particular that join-semilattices that are obtained from dismantable lattices by removing the bottom element are winnable; these are lattices that can be emptied by recursively removing the doubly irreducible elements, see 
Kelly, David; Rival, Ivan, Crowns, fences, and dismantlable lattices, Can. J. Math. 26, 1257-1271 (1974). This paper shows in particular that dismantable lattices are exactly the lattices that are crown-free (see the paper for the definition), so if we want to solve our problem for join-semilattices only (instead of for all posets), it becomes equivalent to the following:

Can we win for every join-semilattice that has a crown? (Note that the second example in the question is a join-semilattice that has a crown).

Other than claim (*), we have some code (there) to decide if a given poset is winnable or not, and if yes to visualize the winning strategy. In particular we have generated a lot of join-semilattices and found no counter-example so far.

We are also working on a generalization of this problem: see this note.