In my question, I denote by $|\cdot|$ the cardinality of any set. Moreover, if $f: X \to Y$, we denote by $\mathcal{P}f$ its direct image, i.e. $\mathcal{P}f(A)=\{f(a) : a \in A\}$. Let $X,Y$ be two sets.
Is it true that $|X| \le |Y| \iff |\mathcal{P}(X)| \le |\mathcal{P}(Y)|$?
My attempt. Assume first that $|X| \le |Y|$. Then there exists an injective function $f: X \to Y$. Hence we may easily check that the direct image $\mathcal{P}f: \mathcal{P}(X) \to \mathcal{P}(Y)$ is also injective, so that $|\mathcal{P}(X)| \le |\mathcal{P}(Y)|$.
[Actually, we should have $|X| < |Y| \implies |\mathcal{P}(X)| < |\mathcal{P}(Y)|$ because if the direct image $\mathcal{P}g$ of some function $g: X \to Y$ is surjective, then also $g$ is a surjection]
Conversely, assume that $|\mathcal{P}(X)| \le |\mathcal{P}(Y)|$ and suppose by contradiction that $|X|>|Y|$. Then by the above step we should have $|\mathcal{P}(X)| > |\mathcal{P}(Y)|$, absurd.
Is my argument correct? Furthermore, is it also true that $|\mathcal{P}(X)| < |\mathcal{P}(Y)| \implies |X| < |Y|$?
Your first argument (showing that $|X|\leq |Y| \implies|\mathcal{P}(X)|\leq |\mathcal{P}(Y)|$) is correct. (That is, assuming that your proof for the injectivity of the direct image function $\mathcal{P}f$ goes through $-$ this statement is, in any case, correct.)
Everything that follows, alas, is not. In fact, the reverse implication is independent from ZFC. By Easton's theorem, it is consistent with ZFC that, say, $|\mathcal{P}(\aleph_0)|=|\mathcal{P}(\aleph_1)|$, or even that all powersets of all infinite cardinals up to $\aleph_{\aleph_1}$ or anything of the like, have the same size.
As for your last question: That implication is just the contrapositive of the first, so it is true as well.