Theorem 1. If $V$ is a non-trivial set generic extension of $W\models\mathrm{ZFC}$ then there is no
$j:V\to W$ as described (i.e. with $x\in y\iff j(x)\in j(y)$ for all $x,y\in V$).
(So in particular, regarding a question at the end of @JoelDavidHamkins' answer, if $V=L[c]$ where $c$ is Cohen generic over $L$, then there is no such $j:V\to L$.)
Proof. Suppose otherwise and let
$\mathbb{P}\subseteq\alpha\in\mathrm{OR}$ with $\mathbb{P}\in W$ and
$G$ be $(W,\mathbb{P})$-generic with $G\notin W$. Let $j:V\to W$ be the embedding. I'll argue somewhat
like in the proof of Theorem 2.3/Lemma 2.2 of Schlutzenberg - Reinhardt cardinals and iterates of $V$ for a contradiction.
Let $\beta$ be a regular cardinal $>\alpha$.
Let $\dot{k}\in W$ be a $\mathbb{P}$-name for $j\upharpoonright\beta$. Working in
$W$, by using the $\mathbb{P}$-forcing relation, we can find some set
$A\subseteq\beta$ of ordertype $\alpha$ and some $p\in G$ such that $p$
decides the value of $\dot{k}(\gamma)$ for each $\gamma\in A$.
Therefore $j\upharpoonright A\in W$. Working in $V$, let $G'\subseteq
A$ be the "translation" of $G$; that is, let $\pi:\alpha\to A$ be the increasing enumeration
of $A$, and let $G'=\pi``G$. Let $G^*=j(G)\in W$. Then working in $W$, from $G^*$ and $j\upharpoonright A$, we can compute $G'$ (i.e. for $\gamma\in A$, we have $\gamma\in G'$ iff $j(\gamma)\in G^*$; note this uses only the elementarity of $j$ that is assumed). But $A,\pi\in W$ also, and from these and $G'$ we can compute $G$, so $G\in W$, a contradiction.
Remark. By Theorem 2.3 of the paper mentioned above, if $W\models\mathrm{ZF}$ and $V$ is a set-generic extension of $W$ and $j:V\to W$ is elementary, then $W,V$ have the same sets of ordinals. By adapting the foregoing argument with that for Theorem 2.3, the elementarity assumption of Theorem 2.3 can be reduced to the weak elementarity considered here.
(Update) Theorem 2. Suppose ZFC + $j:V\to L$ is an embedding. Then CH holds, and therefore by the earlier answers and comments, GCH holds.
Proof. This is a variant of the proof of GCH $>\aleph_0$ from earlier answers and comments. For $i=0,1$, let $\eta_{\aleph_i}$ be the least ordinal $\eta$ such that there is a map $\pi:\aleph_i\to\eta$ and a map $\pi^+:\mathcal{P}(\aleph_i)\to\mathcal{P}(\eta)^L$ such that for all $\alpha<\aleph_i$ and all $X\subseteq\aleph_i$, we have $\pi(\alpha)\in\pi^+(X)$ iff $\alpha\in X$.
(Correction: I asserted in an earlier version of this that we get such a $\pi,\pi^+$ by restricting $j$, but that was confused, since $j$ need not map ordinals to ordinals. But we can easily modify $j$ to get such a $\pi,\pi^+$. That is, let $\sigma\in L$ be a bijection $\sigma:j(\aleph_i)\to\eta$ where $\eta\in\mathrm{OR}$. Let $\pi=\sigma\circ j\upharpoonright\aleph_i$. For $X\subseteq\aleph_i$ let $\pi^+(X)=\sigma``(j(X)\cap j(\aleph_i))$, noting $\pi^+(X)\in L$. Then $\eta,\pi,\pi^+$ work.)
Claim 1. $\eta_{\aleph_i}$ is an $L$-cardinal, for $i=0,1$.
Proof. Suppose not and let $\eta=\mathrm{card}^L(\eta_{\aleph_i})$. Let $\sigma:\eta\to\eta_{\aleph_i}$ be a bijection with $\sigma\in L$. Let $\pi,\pi^+$ witness the definition of $\eta_{\aleph_i}$. Then using $\pi,\pi^+,\sigma$ we can construct a witness to show that $\eta_{\aleph_i}\leq\eta$, a contradiction.
(That is, define $\pi'(\alpha)=\sigma^{-1}(\pi(\alpha))$,
and define $(\pi^+)'(X)=\sigma^{-1}``\pi(X)$, noting that $(\pi^+)'(X)\in L$.)
Claim 2. $\eta_{\aleph_0}<\eta_{\aleph_1}$.
Proof. Easily $\eta_{\aleph_0}\leq\eta_{\aleph_1}$. So (let and) suppose $\eta=\eta_{\aleph_0}=\eta_{\aleph_1}$. Note that $\mathrm{cof}(\eta)=\aleph_0$ (this is $V$-cofinality), as $\eta=\eta_{\aleph_0}$. Let $\pi,\pi^+$ witness the definition of $\eta_{\aleph_1}=\eta$. Then we can fix $\eta'<\eta$ such that $\mathrm{rg}(\pi)\cap\eta'$ has cardinality $\aleph_1$. Let $A=\pi^{-1}``\eta'$. Let $\sigma:A\to\eta'$ be $\sigma=\pi\upharpoonright A$ and let $\sigma^+:\mathcal{P}(A)\to\mathcal{P}(\eta')^L$ be $\sigma^+(X)=\pi^+(X)\cap\eta'$ (note $\sigma^+(X)\in L$). Now shift $\sigma,\sigma^+$ to have domain $\aleph_1$ instead (but with the same range etc). This shows that $\eta_{\aleph_1}\leq\eta'$, a contradiction.
Now if $\eta_{\aleph_0}<\aleph_1$ then as in the proof of GCH above $\aleph_0$,
we get $2^{\aleph_0}=\aleph_1$. So suppose $\eta=\eta_{\aleph_0}\geq\aleph_1$. We have $\eta_{\aleph_1}<\aleph_2$, by the proof of GCH above $\aleph_0$ (that is, $j``\aleph_1$ is covered by some set $B\in L$ of ($V$-)cardinality $\aleph_1$, and working in $L$, we can shift $B$ down to its ordertype $\eta'$, and $\eta_{\aleph_1}\leq\eta'<\aleph_2$).
So all together and by the claims,
$$\aleph_1<\eta=\eta_{\aleph_0}<\eta^{+L}\leq\eta_{\aleph_1}<\aleph_2.$$
But then again as in the proof of GCH above $\aleph_0$, we can embed $\mathcal{P}(\omega)$ injectively into $\eta^{+L}$, and therefore $2^{\aleph_0}=\aleph_1$, as desired.
(Update 2) By similar reasoning we can also get a strong form of GCH at uncountable cardinals:
Theorem 3. Assume ZFC + $j:V\to L$ is an embedding where $V\neq L$.
Then for every uncountable cardinal $\kappa$, there is a set $A\subseteq\kappa$ such that $\mathcal{P}(\kappa)\subseteq L[A]$ (and hence $\mathcal{P}(\kappa)\subseteq L_{\kappa^+}[A]$).
Proof. Define $\eta_\kappa$ for $\kappa$ just like $\eta_{\aleph_1}$ was defined above. By covering, $\kappa\leq\eta_\kappa<\kappa^+$, and $\eta_\kappa$ is an $L$-cardinal, like before. Moreover, note that $(\eta_\kappa)^{+L}=\kappa^+$;
in other words, $\eta_\kappa$ is the largest $L$-cardinal which is ${<\kappa^+}$ (and either $\kappa$ is regular and $\mathrm{cof}(\eta_\kappa)=\kappa$, or $\kappa$ is singular and $\eta_\kappa=\kappa$ (by covering)). Let $\pi:\kappa\to\eta_\kappa$ and $\pi^+$ witness the definition of $\eta_\kappa$. Then $\mathcal{P}(\kappa)\subseteq L_{\kappa^+}[\pi]$ (for note that for each $X\subseteq\kappa$, we can compute $X$ from $\pi^+(X)$ and $\pi$, and $\pi^+(X)\in L_{\kappa^+}$; we don't need $\pi^+$ itself to do this). But $\pi$ is coded by some $A\subseteq\kappa$, so we are done.
(Of course if $\eta_{\aleph_0}<\aleph_1$ then we also get $\mathcal{P}(\omega)\subseteq L[x]$ for some real $x$. But I don't see that $\eta_{\aleph_0}<\aleph_1$, and if $\eta_{\aleph_0}>\aleph_1$ then I don't see why there should be such an $x$.)
Hamkins proved already under ZFC + "$j:V\to L$ is an embedding" that $0^\#$ does not exist. This can be refined as follows:
(Update 3) Theorem 4. Assume ZFC + $j:\mathcal{P}(\omega)\to L$ is an embedding. Then $0^\#$ does not exist.
Proof. Suppose otherwise. For each $n<\omega$ let $t_n$ be a term and $\vec{\kappa}_n$ a finite tuple of Silver indiscernibles such that $j(n)=t_n^L(\vec{\kappa}_n)$. Likewise define $t_x$ and $\vec{\kappa}_x$ for $x\subseteq\omega$. Then for $x\neq y$ we have $(t_x,\vec{\kappa}_x)\neq(t_y,\vec{\kappa}_y)$. But because $\mathcal{P}(\omega)$ is uncountable, there will be $x\neq y$ such that $t_x=t_y$ and $\vec{\kappa}_x,\vec{\kappa}_y$ have the same "type" with respect to $\left<\vec{\kappa}_n\right>_{n<\omega}$;
that is, $\vec{\kappa}_x$ sits in $\vec{\kappa}_x\cup\bigcup_{n<\omega}\vec{\kappa}_n$ in terms of ordertype position just as $\vec{\kappa}_y$ sits in $\vec{\kappa}_y\cup\bigcup_{n<\omega}\vec{\kappa}_n$. (That is, consider the function $x\mapsto(t_x,\mathrm{type}(\vec{\kappa}_x))$ in this sense;
there are only countably many values in the range, so we get $x\neq y$ with the same output.) But then by indiscernibility, it follows that $j(x)$ and $j(y)$ agree with each other on membership with regard to $j(n)$ for each $n<\omega$, which contradicts that $j$ is an embedding.
