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Exclusive or ("one or the other, and not both") is the same as inclusive or ("one or the other, or both") when the operands are not both true.

Consider the truth tables:

| $A$ | $B$ | $A \oplus B$ | $A \lor B$ | $A \oplus B = A \lor B$ |
|:---:|:---:|:------------:|:----------:|:-----------------------:|
| 0 | 0 | 0 | 0 | 1 |
| 0 | 1 | 1 | 1 | 1 |
| 1 | 0 | 1 | 1 | 1 |
| 1 | 1 | 1 | 0 | 0 |

From this, we see that $A \oplus B = A \lor B \iff ¬(A \land B)$. We can test this condition against your two examples:

  * \begin{align}¬(X \land ¬X) &\iff ¬0 \\&\iff 1\end{align}
  * \begin{align}¬((A \land ¬B) \land (¬A \land C)) &\iff ¬(A \land ¬B \land ¬A \land C) \\&\iff ¬((A \land ¬A) \land ¬B \land C) \\&\iff ¬(0 \land ¬B \land C) \\&\iff ¬0 \\&\iff 1\end{align}

Therefore, the simplification is valid in both cases.

As observed [by Olin Lathrop](#answer-295343), we can perform the same calculation with other operations, to find the conditions under which they can be simplified to each other. This is not a particularly _sophisticated_ trick, but it can be useful if you wish to avoid exhaustive case analysis.