First: terminology. An argument is a collection of statements (more details below) while a tautology or contradiction is a single statement. So with regard to your last question, it makes no sense to say that "an argument is equivalent to a contradiction" - they are different kinds of objects.
BTW a remark on the side - one of the first things you need to do in studying any mathematical topic is to make sure you know what kinds of things you are talking about. For example, I frequently have students attempting a problem concerning a function $T$ and trying to prove that $0$ is an element of $T$. This makes no sense at all because a function does not have "elements", only a set can have elements.
Second: you shouldn't go overboard on notation. It's very easy to forget that the symbols actually mean something and just try to apply rules to them. However in your question you have used no notation at all and I think some would be helpful. I will assume that you are studying propositional logic; if you later move on to predicate logic, similar ideas apply.
So, an individual statement of propositional logic is a tautology if it is always true, and a contradiction if it is always false. A very simple example of a tautology is
$$p\vee(\neg p)\ ;$$
informally this means "$p$ is either true or false", and this statement is clearly true as a whole, regardless of whether $p$ is in fact true or false. Likewise,
$$p\wedge(\neg p)$$
is a contradiction since it is inevitably false, no matter what $p$ may be. Just a very simple example like
$$p$$
could be true and could be false, so it is neither a tautology nor a contradiction. For a harder example such as
$$(p\to(q\vee\neg r))\to(r\to(p\to q))\tag{$*$}$$
you would draw up a truth table and observe whether the final result is all true, all false or neither of these. I'll leave this one for you as an exercise.
Next: an argument is a (finite) sequence of statements. The last statement in the sequence is called the conclusion and the preceding ones are the premises or assumptions or hypotheses. Saying that an argument is valid essentially means that the logic is correct, regardless of the actual facts asserted: more precisely, it means that in any case where all the hypotheses are true, the conclusion must also be true. The best known example is
$$\eqalign{&p\to q\cr &p\cr \therefore\ &q\ ,\cr}$$
sometimes referred to as modus ponens. Hopefully it is intuitively clear that if $p\to q$ and $p$ are true then $q$ must be true: essentially, this is just the meaning of "if. . . then". Another (slightly silly) example is
$$\eqalign{&p\cr &\neg p\cr \therefore\ &p\wedge(\neg p)\ .\cr}$$
Of course in this case the two hypotheses cannot be simultaneously true; but it is obvious that if they were then the conclusion would have to be true: this is really no more than the meaning of the word "and". So the argument is valid, even though the conclusion is false! Here is an example of an invalid argument:
$$\eqalign{&p\to(q\vee r)\cr &q\to r\cr &\neg q\cr \therefore\ &\neg p\ .\cr}$$
You can find by trial and error that if $p$ is true, $q$ false and $r$ true, then the three assumptions are true but the conclusion is false. Now suppose that we have an argument
$$\eqalign{&P_1\cr &\vdots\cr &P_n\cr \therefore\ &Q\ .\cr}$$
To say that this is a valid argument means
in all cases, if $P_1,\ldots,P_n$ are all true then $Q$ is true;
that is,
in all cases, if $P_1\wedge\cdots\wedge P_n$ is true, then $Q$ is true;
that is,
in all cases, $(P_1\wedge\cdots\wedge P_n)\to Q$ is true.
That is, the above argument is valid if and only if the statement
$$(P_1\wedge\cdots\wedge P_n)\to Q$$
is a tautology. So, if you have an argument where the trial and error is too hard, you can construct this statement, write up a truth table, decide whether the statement is a tautology or not, and hence decide whether the argument is valid or not.
The case you are asking for, where the truth values turn out to be always false, is not really of any particular interest - the idea is that for a valid argument, if we confirm that the premises are true, then we can be certain that the conclusion is true. So if the truth values are even sometimes false, the argument is of no value. However, it is certainly possible to construct an "extreme" case: you would need $P_1,\ldots,P_n$ to be always true, that is, tautologies, and $Q$ always false, that is, a contradiction. An example would be
$$\eqalign{&p\vee(\neg p)\cr &p\to p\cr \therefore\ &p\wedge(\neg p)\ .\cr}$$
Hope this helps. Good luck!
Answer to $(*)$: it's a tautology.
A proof by contradiction has an important step that you omitted: you start from a false premise, reason to a contradiction, and then discard the false premise as incorrect. It is in fact exactly the avoidance of "proof by explosion."
In other words, we assume various proven facts, call them $A,B,C...$, as well as facing a claim that $P \implies Q$. If you start from an assumption of $\neg (P \implies Q)$, and use it to derive $\neg C$, you have shown that $\neg (P \implies Q) \implies \neg C$, and the contrapositive of that is $C \implies (P \implies Q)$. Since you know $C$ is true, you can conclude that $P \implies Q$.
Best Answer
Also: $\quad(\lnot p{\kern.6em\not\kern-.6em\implies}\bot)\;\equiv\;(\lnot p\land\lnot \bot)\;\equiv\;(\lnot p\land\top)\;\equiv\;\lnot p.$
This requires establishing that the antecedent $¬p$ is true (since a false antecedent automatically makes the implication true).
This also requires establishing that the antecedent $¬p$ is true (since a false antecedent can always lead to a contradiction).
But $¬p$ being true directly means that $p$ is false, without going through the above detours.
This does not entail that $p$ is false; for example: $¬(A\lor\lnot A)\implies\top.$
Summary: your proposal to disprove $p$ by showing that $\lnot p$ leads to no contradiction is not illogical—merely a circuitous way of disproving $p$ by proving $\lnot p.$