There's still one missing bit of information, and that's the probability that the witch would save a dying player other than herself (we're assuming that she would always save herself on the first night if she needed to).
For now, let's assume that she's very selfish and would never use her potion on anyone other than herself. We can come back to this assumption later.
First, what are the odds that no one dies after the first night? Let's call this $P(s)$, for "probability of survival". Under our selfish-witch assumption, survival can only happen if the werewolves tried to attack a player under the protection of the guardian angel, or if they attacked the witch (who would save herself). Let's call the probability of the former happening $P(g)$, for "probability that the attacked player was guarded", and the latter $P(w)$, for "probability that the attacked player was the witch". So, we get the following:
$$
P(s) = P(g) + P(w) - P(g \cap w)
$$
The last term represents the case where the werewolves attack the witch, who is also the one being protected by the guardian angel. The reason we need to subtract this case out is because it would otherwise be counted twice! Once in $P(g)$ and once in $P(w)$. Read up about the inclusion-exclusion principle for more detail on this.
You calculated $P(g)$ correctly as the probability that the guardian angel picked a non-werewolf times the probability that the werewolves attacked that person. That is:
$$
P(g) = \frac{7}{10}\times\frac{1}{8} = \frac{7}{80} = 0.0875
$$
The probability that the werewolves attacked the witch is simply:
$$
P(w) = \frac{1}{8} = 0.125
$$
The odds that the werewolves attack a guardian-angel-protected witch can be calculated by multiplying together the probability that the guardian angel decided to protect the witch (1 in 10) and the probability that the werewolves decided to attack the doubly protected player (1 in 8). In other words:
$$
P(g \cap w) = \frac{1}{10}\times\frac{1}{8} = \frac{1}{80} = 0.0125
$$
So, if we plug these into our equation for $P(s)$, we get:
$$
P(s) = 0.0875 + 0.125 - 0.0125 = 0.2
$$
As Greg Snow mentioned, we can use Bayes' theorem to find our answer. Since we want to know how likely it is that the guardian angel is the reason for a first night where no one dies, what we're trying to solve for looks like this: $P(g|s)$. This is read as "the probability of $g$ given that $s$ is true", or just "the probability of $g$ given $s$". As Bayes' theorem states, this is equal to the following:
$$
P(g|s) = \frac{P(g)\times P(s|g)}{P(s)}
$$
We know that $P(g)$ is $0.0875$, and we know that $P(s)$ is $0.2$, but what is $P(s| g)$? That is, what is the probability that no one dies the first night given that the werewolves attacked a guardian-angel-protected player? Clearly, that is always $1$. So:
$$
P(g|s) = \frac{0.0875 \times 1}{0.2} = \frac{0.0875}{0.2} = 0.4375
$$
And there you go! Assuming the witch always saves her potion for herself, the likelihood that a lucky guess by the guardian angel caused the safe night is $43.75\%$.
If we know that the witch may be willing to part with her potion a particular percentage of the time, we can simply update our $P(s)$. Let's let this probability of compassion be called $P(c)$. Then:
$$
P(s) = 0.2 \times 1 + 0.8 \times P(c)
$$
That is, there will always be a safe night if either the witch or the protected player is attacked (with chance of $0.2$) and a safe night with probability $P(c)$ otherwise (with chance of $1.0 - 0.2 = 0.8$). Updating our $P(g|s)$, we get:
$$
P(g|s) = \frac{0.0875}{\left(0.2 + 0.8 \times P(c)\right)}
$$
Sorry if that was difficult to read because of poor formatting or if it was too long-winded. This is also my first answer on this Stack Exchange board. :P
I think you are making a mistake in the denominator of the former.
$P(A) = \sum_B P(A|B) P(B)$
Since rushing made me write something kinda terrible I'll pay some penance by writing out a more full solution!
The Bayes Approach to the Monty Hall Problem
In general I think you are making the Monty Hall problem a little bit more confusing when you omit the player's choice. The player chooses an arbitrary door so without loss of generality let's call it one. I believe this is what you did, but let's just be a bit more explicit.
Next we write out the probability:
$P(H=1|D=3) = \frac{P(H=1) P(D=3|H=1)}{P(D=3)} $
We can calculate the numerator as $P(H=1) = \frac{1}{3}$ since each door is equally likely. We also calculate $P(D=3|H=1) = \frac{1}{2}$ since if $H=1$ Monty can pick either door without showing the grand prize. Moving on:
$P(H=1|D=3) = \frac{\frac{1}{3} \frac{1}{2}}{P(D=3|H=1)*P(H=1)+P(D=3|H=2)P(H=2)+P(D=3|H=3)P(H=3)}$
Next we have a set of terms to consider: $P(H=i)$ is $\frac{1}{3}$ for the same reason as above. Finally the conditionals: $P(D=3|H=1)=\frac{1}{2}$ as above. $P(D=3|H=2) = 1$ this is because Monty can't show door 1 since the player picked it, Monty can't show door 2 because it has the grand prize, and thus Monty must show door 3. Finally there is $P(D=3|H=3) = 0$ this is because Monty can't show the prize. Thus:
$P(H=1|D=3)= \frac{\frac{1}{3} \frac{1}{2}}{(\frac{1}{2}+ 1+ 0)(1/3)} = \frac{\frac{1}{3}\frac{1}{2}}{\frac{1}{3} \frac{3}{2}}=1/3$
Lonely Monty Hall Problem
This is not to be interpreted as the standard Monty Hall problem. Say that we revisit the problem without a player. There are again three doors, one of which has a grand prize behind it. Monty will choose and open a door that doesn't have the grand prize behind it and we are tasked with evaluating the probability that the grand prize is behind door 1.
The original write up of the question doesn't explicitly include the player so this is a fair interpretation.
Calculation via Bayes Theorem:
Let's start by just writing out the expression:
$P(H=1|D=3) = \frac{P(H=1) P(D=3|H=1)}{P(D=3)}$
Working from here $P(H=1) = \frac{1}{3}$ and $P(D=3|H=1)=\frac{1}{2}$ as above. We would then expand the denominator:
$P(D=3) = \sum_i P(D=3|H=i) P(H=i)$
Now what's different is that the player has not made a choice on door so we need to calculate the various conditional probabilities. $P(D=3|H=1) = \frac{1}{2}$ since Monty has two choices open door 2 or door 3. Similarly, $P(D=3|H=2) = \frac{1}{2}$, finally $P(D=3|H=3) = 0$ since Monty can not reveal the prize.
Overall this gives:
$P(H=1|D=3) = \frac{\frac{1}{3} \frac{1}{2}}{\frac{1}{3} \frac{1}{2}+\frac{1}{3} \frac{1}{2}+\frac{1}{3} 0} = \frac{1}{2}$
Best Answer
Let's start with the regular Monty Hall problem. Three doors, behind one of which is a car. The other two have goats behind them. You pick door number 1 and Monty opens door number 2 to show you there is a goat behind that one. Should you switch your guess to door number 3? (Note that the numbers we use to refer to each door don't matter here. We could choose any order and the problem is the same, so to simplify things we can just use this numbering.)
The answer of course is yes, as you already know, but let's go through the calculations to see how they change later. Let $C$ be the index of the door with the car and $M$ denote the event that Monty revealed that door 2 has a goat. We need to calculate $p(C=3|M)$. If this is larger than $1/2$, we need to switch our guess to that door (since we only have two remaining options). This probability is given by: $$ p(C=3|M)=\frac{p(M|C=3)}{p(M|C=1)+p(M|C=2)+p(M|C=3)} $$ (This is just applying Bayes' rule with a flat prior on $C$.) $p(M|C=3)$ equals 1: if the car is behind door number 3 then Monty had no choice but to open door number 2 as he did. $p(M|C=1)$ equals $1/2$: if the car is behind door 1, then Monty had a choice of opening either one of the remaining doors, 2 or 3. $p(M|C=2)$ equals 0, because Monty never opens the door that he knows has the car. Filling in these numbers, we get: $$ p(C=3|M)=\frac{1}{0.5+0+1}=\frac{2}{3} $$ Which is the result we're familiar with.
Now let's consider the case where Monty doesn't have perfect knowledge of which door has the car. So, when he chooses his door (which we'll keep referring to as door number 2), he might accidentally choose the one with the car, because he thinks it has a goat. Let $C'$ be the door that Monty thinks has the car, and let $p(C'|C)$ be the probability that he thinks the car is in a certain place, conditional on its actual location. We'll assume that this is described by a single parameter $q$ that determines his accuracy, such that: $p(C'=x|C=x) = q = 1-p(C'\neq x|C=x)$. If $q$ equals 1, Monty is always right. If $q$ is 0, Monty is always wrong (which is still informative). If $q$ is $1/3$, Monty's information is no better than random guessing.
This means that we now have: $$p(M|C=3) = \sum_x p(M|C'=x)p(C'=x|C=3)$$ $$= p(M|C'=1)p(C'=1|C=3) + p(M|C'=2)p(C'=2|C=3) + p(M|C'=3)p(C'=3|C=3)$$ $$= \frac{1}{2} \times \frac{1}{2}(1-q) + 0\times \frac{1}{2}(1-q) + 1 \times q$$ $$= \frac{1}{4} - \frac{q}{4} + q = \frac{3}{4}q+\frac{1}{4}$$
That is, if the car was truly behind door 3, there were three possibilities that could have played out: (1) Monty thought it was behind 1, (2) Monty thought 2 or (3) Monty thought 3. The last option occurs with probability $q$ (how often he gets it right), the other two split the probability that he gets it wrong $(1-q)$ between them. Then, given each scenario, what's the probability that he would have chosen to point at door number 2, as he did? If he thought the car was behind 1, that probability was 1 in 2, as he could have chosen 2 or 3. If he thought it was behind 2, he would have never chosen to point at 2. If he thought it was behind 3, he would always have chosen 2.
We can similarly work out the remaining probabilities: $$p(M|C=1) = \sum_x p(M|C'=x)p(C'=x|C=1)$$ $$=\frac{1}{2}\times q + 1\times \frac{1}{2}(1-q)$$ $$=\frac{q}{2}+\frac{1}{2}-\frac{q}{2}=\frac{1}{2}$$
$$p(M|C=2) = \sum_x p(M|C'=x)p(C'=x|C=2)$$ $$=\frac{1}{2}\times\frac{1}{2}(1-q) + 1 \times\frac{1}{2}(1-q)$$ $$=\frac{3}{4}-\frac{3}{4}q$$
Filling this all in, we get: $$ p(C=3|M)=\frac{\frac{3}{4}q+\frac{1}{4}}{\frac{1}{2}+\frac{3}{4}-\frac{3}{4}q+\frac{3}{4}q+\frac{1}{4}} $$ $$ =\frac{0.75q+0.25}{1.5} $$ As a sanity check, when $q=1$, we can see that we get back our original answer of $\frac{1}{1.5}=\frac{2}{3}$.
So, when should we switch? I'll assume for simplicity that we're not allowed to switch to the door Monty pointed to. And in fact, as long as Monty is at least somewhat likely to be correct (more so than random guessing), the door he points to will always be less likely than the others to have the car, so this isn't a viable option for us anyway. So we only need to consider the probabilities of doors 1 and 3. But whereas it used to be impossible for the car to be behind door 2, this option now has non-zero probability, and so it's no longer the case that we should switch when $p(C=3|M)>0.5$, but rather we should switch when $p(C=3|M)>p(C=1|M)$ (which used to be the same thing). This probability is given by $p(C=1|M)=\frac{0.5}{1.5}=\frac{1}{3}$, same as in the original Monty Hall problem. (This makes sense since Monty can never point towards door 1, regardless of what's behind it, and so he cannot provide information about that door. Rather, when his accuracy drops below 100%, the effect is that some probability "leaks" towards door 2 actually having the car.) So, we need to find $q$ such that $p(C=3|M) > \frac{1}{3}$: $$\frac{0.75q+0.25}{1.5}>\frac{1}{3}$$ $$0.75q+0.25 > 0.5$$ $$0.75q > 0.25$$ $$q > \frac{1}{3}$$ So basically, this was a very long-winded way to find out that, as long as Monty's knowledge about the car's true location is better than a random guess, you should switch doors (which is actually kind of obvious, when you think about it). We can also calculate how much more likely we are to win when we switch, as a function of Monty's accuracy, as this is given by: $$\frac{p(C=3|M)}{p(C=1|M)}$$ $$=\frac{\frac{0.75q+0.25}{1.5}}{\frac{1}{3}}=1.5q+0.5$$ (Which, when $q=1$, gives an answer of 2, matching the fact that we double our chances of winning by switching doors in the original Monty Hall problem.)
Edit: People were asking about the scenario where we are allowed to switch to the door that Monty points to, which becomes advantageous when $q<\frac{1}{3}$, i.e. when Monty is a (somewhat) reliable "liar". In the most extreme scenario, when $q=0$, this means the door Monty thinks has the car actually for sure has a goat. Note, though, that the remaining two doors could still have either a car or a goat.
The benefit of switching to door 2 is given by: $$ \frac{p(C=2|M)}{p(C=1|M)} = \frac{ \frac{0.75 - 0.75q}{1.5} } { \frac{1}{3} } = 1.5-1.5q $$ Which is only larger than 1 (and thus worth switching to that door) if $1.5q<0.5$, i.e. if $q<\frac{1}{3}$, which we already established was the tipping point. Interestingly, the maximum possible benefit for switching to door 2, when $q=0$, is only 1.5, as compared to a doubling of your winning odds in the original Monty Hall problem (when $q=1$).
The general solution is given by combining these two switching strategies: when $q>\frac{1}{3}$, you always switch to door 3; otherwise, switch to door 2.