What is the probability of drawing two cards without replacement?

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Two events $A$ and $B$ are independent if & only if $\Pr(A \cap B) = \Pr(A)\Pr(B)$.

You can divide both sides of this equation by $\Pr(A)$, to get $$ \frac{\Pr(A \cap B)}{\Pr(A)} = \Pr(B) $$ Notice now that the term on the left is just the definition of $\Pr(B|A)$. So we'e shown that $A$ and $B$ are independent if & only if $\Pr(B|A) = \Pr(B)$.

Let's take your definitions of $A$ as the event that the 1st card is an Ace, and $B$ as the event that the 2nd card is a Spade. The probability of the 2nd card being a Spade is $\tfrac{12}{51}$ in case the first card was also a Spade, and $\tfrac{13}{51}$ if it was not. So $$ \Pr(B) = \frac{1}{4}\cdot\frac{12}{51} + \frac{3}{4}\cdot\frac{13}{51} = \frac{51}{4\cdot 51} = \frac{1}{4} $$ Let us now consider $\Pr(B|A)$. The probability of the 2nd card being a Spade is again $\tfrac{12}{51}$ in case the first card was the Ace of Spades, and $\tfrac{13}{51}$ if it was one of the other 3 aces. This is the exact same: $\Pr(B|A)=\Pr(B)=\tfrac{1}{4}$, hence the events are indeed independent.

I think the mistake in your reasoning was where you said, "$\Pr(B|A)=\tfrac{12}{51}$ if $A$ is the Ace of Spades, but $\Pr(B|A)=\tfrac{13}{51}$ if $A$ is not the Ace of Spades". The problem with this rationale is that $\Pr(B|A)$ cannot have two different values, and $A$ is an event, not a card. $\Pr(B|A)$ has a single composite value, which is made up from the $\tfrac{12}{51}$ and $\tfrac{13}{51}$ probabilities you mentioned, but weighted by the (respective) $\tfrac{1}{4}$ and $\tfrac{3}{4}$ probabilities of the 1st Ace being the Ace of Spades vs. not (or, of event $A$ being True or False).

As an aside, what if we let $C$ be the event that the 1st card is a Black card? The probability of drawing a Spade, after we have already drawn a Black card, is $\tfrac{12}{51}$ if that Black card was also a Spade, and $\tfrac{13}{51}$ if it was a Club: $$ \Pr(B|C) = \frac{1}{2}\cdot\frac{12}{51} + \frac{1}{2}\cdot\frac{13}{51} = \frac{25}{102} \approx 0.245 \neq 0.25 = \frac{1}{4} = \Pr(B) $$ In this case, $B$ and $C$ are not independent.

EDIT: Sure, even if $C$ is true, we still don't know whether the 1st card was a Spade, or not. But because all Spades are Black (although not the other way around), knowing that the 1st card was already Black, gives us some additional relevant information, about the chance that the 2nd card will be a Spade. In contrast, knowing that the 1st card was an Ace isn't really relevant to the question of whether the 2nd card will be a Spade, because Spades are $\tfrac{1}{4}$ of all Aces, but Spades are also $\tfrac{1}{4}$ of the deck as a whole.

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