When "If $A$ is true then $B$ is true", is it valid to assert that "If $B$ is false, $A$ must also be false"? If it is given that: 

"People that ride buses, also ride planes"

then is the statement 

"people that don't ride planes, also don't ride buses"

necessarily true?
I don't think so, but the explanation to a problem in textbook I'm using uses that as logical proof to the answer provided.
It is based on the rule (according to this textbook) that, if this is true : ( if $A$ is true, then $B$ is also true), then it follows that if $B$ is false, then $A$ must also be false.
 A: I think the problem OP has is that the statement given is about propositional(zero-order) logic but to solve his confusion he will need predicate(first-order) logic, i.e. both "people"s in the statement means all people by the author. Because clearly there are some people, in real world, that don't ride planes, but ride buses. You may take a look about predicate logic.
When you say people, you should be clear about what you meant:
People that ride buses, also ride planes: Do you meant all people? If this is the case then you're saying
$$\forall x, P(x)\to Q(x), x=\textrm{people}.$$
Since there are some people in the world that "don't ride planes, but ride buses" but that's your another problem: When someone say
$$\textrm{If}\ A\ \textrm{then}\ B,$$
in your case $A:$ (all) people that ride buses. "$A$" don't have to be true in read world. What this logical statement(assumed true) means is that $B$ must be true when I suppose $A$ true.

Notice that when you interpret $A$ as
$$\exists x, P(x)\to Q(x),$$
in your case $P(people)=$ some people ride buses; $Q(people)=$ they(same people) also ride planes.
Then yes the conclusion
$$\forall x, \lnot Q(x)\to \lnot P(x), (\textrm{which is }\equiv(\forall x, P(x)\to Q(x))),$$
is false because this is stronger then the original one. You implicitly changed $\exists$(exists) to $\forall$(for all) in your brain, but that's fine because when we doubt a thing we will be trying to find the counter example implicitly. That's why we extend the propositional logic to predicate logic, because the latter is more precise.
A: The difference lies in the logical statement 

If $A$ is true, then $B$ is true.

Which is distinct from

(The truth value of) $A$ implies $B$.

You need to go beyond the examples you can provide with colloquial English, since a statement like 

It is raining implies it is cloudy.

Has an exception, since "sun showers" exist. 
Now, you're meant to regard $A \implies B$ as an agreement. The agreement being 

The statement $A \rightarrow B$ is true, and so is the statement $A,$ then we agree that $B$ is true (this is modus ponens).

If you accept this statement, then you can derive the contrapositive: which is that  $A \implies B$ is true precisely when $\neg B \implies \neg A$ is true.
A: What you are asking is does:
A ⟹ B (A implies B)

mean
Not B ⟹ Not A (The converse of B implies the converse of A)?

The answer is yes.
In fact this is the basis of what is known as a Proof By Contradition in Maths.
The way I explain it to my A-Level students (18 year old Mathematicians in the UK) that meet this for the first time is with the following.
Let Statement A = "It is 12th of August 2018"

and
Let Statement B = "It is a Sunday"

Here
A ⟹ B

However, if it is not a Sunday (ie Not B, or the converse of B) is true, then we do not know what date it is with an absolute certainty, but we can say with absolute certainty it is NOT the 12th of August 2018, (ie we can say it is Not A, or we can say it is definitely the converse of statement A). It might be the 11th, it might be the 10th, but it definitely is not the 12th.
The famous one in Maths is the proof of the √2 cannot be expressed exactly as a fraction, ie √2 is irrational. This is done by a Proof by Contradition.
For contradition assume:
 A = "√2 is rational"

If that is true, then
 B = "√2 = a/b, where a and b are whole numbers, and 'a' and 'b' have no factors"

So we assume B is true and with a bit of basic maths we find if √2 = a/b, then both a and b have a factor of 2. So we have establish NOT B.
But from what we have said above:
If A ⟹ B, then Not B ⟹ Not A

So we have established that Not A is true, ie √2 is irrational, ie cannot be expressed as a fraction.
