Your distinction between "true" and "logical value 1" is not one that formal logic generally observes. Here "1" and "true" are synonyms for the same concept.
The meaning of the $\Rightarrow$ connective is what its truth table says it is, neither more nor less -- the truth table defines the connective (in classical logic). Fancy words such as "implication" or "if ... then" are just mnemonics to help you remember what the truth table is, and what the connective is good for -- but when there's a conflict between your intuitive understanding of those words and the truth table, the truth table wins over the words.
The important thing to realize is that $\Rightarrow$ is designed to be used together with a $\forall$. If you try to understand its naked truth table it doesn't seem very motivated -- certainly it can't express any notions of cause and effect, because the truth values of $p$ and $q$ just are what they are in any given world. As long as we're only looking at one possible state of the world, there's not much intuitive meaning in asking "what if $p$ held?" because that implies a wish to consider a world where the truth value of $p$ were different.
The device of standard formal logic that allows us to speak about different worlds is quantifiers. What we want to say is something like
In every possible world where I put in a coin, the machine will spit out a soda.
(though that is a little simplified -- we want to consider a "possible world" to be one where I made a different decision about my coins, not to be one where the machine had inexplicably stopped working even though it does work now. But let's sweep that problem aside for now).
This is the same as saying
In every possible world period, it is true that either I don't put in a coin, or I get a soda.
which logically becomes, using the truth table
For all worlds $x$, the proposition (In world $x$ I put in a coin) $\Rightarrow$ (In world $x$ I get a soda) is true.
Since there's a quantification going on, the truth value of the whole thing is not spoiled by the fact that there are some possible worlds with a broken machine where the $\Rightarrow$ evaluates to true. What interests us is just whether the $\Rightarrow$ evaluates to true every time or not every time. As long as we're in the "not every time" context, the machine is broken, and that conclusion is not affected by the "spurious" local instances of $\Rightarrow$ evaluating to true in particular worlds.
The construction that models (more or less) our intuition about cause and effect (or "if ... then") is not really $\Rightarrow$, but the combination of $\forall\cdots\Rightarrow$.
Unfortunately in the usual style of mathematical prose it is often considered acceptable to leave the quantification implicit, but logically it is there nevertheless. (And to add insult to injury, many systems of formal logic will implicitly treat formulas with free variables as universally quantified too, so even there you get to be sloppy and not call attention to the fact that there's quantification going on.)
Note also that this is the case even in propositional logic where there are no explicit quantifiers at all. To claim that $P\to Q$ is logically valid is to say that in all valuations where $P$ is true, $Q$ will also be true -- there's a quantification built into the meta-logical concept of "logically valid".
First of all, I am confused why you would say that $\neg A \rightarrow B$ should be equivalent to $A \rightarrow B$, given that you just argued that the value of $B$ should not matter, rather than the value of $A$! In fact, when $A$ is true it is no longer the case that the value of $B$ doesn;t matter, and so you immediately get that $\neg A \rightarrow B$ is not the same as $A \rightarrow B$.
What would have made a little more sense is if you would have focused on $A \rightarrow \neg B$ instead, because (as you correctly observed) if $A$ is false, then $A \rightarrow B$ has the same truth-value as $A \rightarrow \neg B$ (namely True). However, that still does not mean that they are equivalent, because equivalence means that they should have the same truth-value under any conditions (and again, you have only shown them to have the same truth-value under the condition that $A$ is False). And so $A \rightarrow \neg B$ is also not equivalent to $A \rightarrow B$
Finally, if you are trying to change the $A$ into a $\neg A$, because $A \rightarrow B$ is true when $A$ is false ... well, that makes even less logical sense. Here is an example to demonstrate your faulty logic. Take statement $\neg A$. This statement is true when $A$ is false ... ok, so by your logic we should be able to change the $A$ with a $\neg A$ and get the same statement? No, because changing the $A$ with a $\neg A$ in $\neg A$ gives us $\neg \neg A$, which is equivalent to just $A$ ... which is not at all equivalent to the original $\neg A$.
Don't confuse statements with their truth-values!!
Best Answer
We do not take implication as some form of causality, but as a form of logical consequence, i.e., that the truth of a first thing forces us to accept that a second thing is true. The essence of "and" is: that one thing and another are true necessitates that each of these two things is true. So with $C\land D$ representing the statement that $C$ and $D$ are true forces us to accept that $C$ is true (it also forces us to accept that $D$ is true). If you do not agree, you may have an unusual notion of "and" (or of implication). And this fact, i.e., that the truth of $C\land D$ forces us to accept that $C$ is true. The accepting-force behind this argument is always correct, no matter whether $C\land D$ itself is actually true. This now precisely what $(C\land D)\Rightarrow C$ symbolizes.