Your solution is almost there. Right now, the advantage of $\mathcal{B}$ is at least
$$
\mathsf{Adv}^\Pi(\mathcal{B}) = \epsilon - \Pr[\mathsf{E}]
$$
where $\epsilon = \mathsf{Adv}^{\Pi'}(\mathcal{A})$ and $\mathsf{E}$ is the event that $\mathcal{A}$ queries $H(sk)$. Of course, this lower bound isn't good enough, because $\mathsf{E}$ might somehow be very likely. Fortunately, $\mathcal{B}$ can actually benefit from $\mathsf{E}$ happening, and turn it into a win.
It does this with some polynomial loss. Let $\mathcal{B}$ log all of $\mathcal{A}$'s oracle queries. For each query $q$, let $\mathcal{B}$ attempt to decrypt the ciphertexts $c_0,c_1$ using $q$ as the secret key. If it gets back $m_b,m_{1-b}$, then $q$ might be the secret key. So $\mathcal{B}$ puts $b$ in its set of potential answers $S$. Finally, $\mathcal{A}$ returns its own guess $b'$, which $\mathcal{B}$ will also put in $S$. In the final step, $\mathcal{B}$ returns a uniformly selected element from $S$.
With this modification, if $\mathsf{E}$ happens, then $\mathcal{B}$ will win with probability at least $1/|S|$ (that is, the likelihood that $\mathcal{B}$ selects a $b$, times the likelihood that that $b$ came from the $\mathsf{E}$ event). Similarly, the likelihood that $\mathcal{B}$ chooses $b'$ and it's correct is at least $\epsilon/|S|$, independent of $\mathsf{E}$.
Thus, with the modification, by the law of total probability,
\begin{align*}
\Pr[\mathcal{B} \text{ wins}]
&=
\Pr[\mathsf{E}] \cdot \Pr[\mathcal{B} \text{ wins} \mid \mathsf{E}]
+ \Pr[\lnot\mathsf{E}] \cdot \Pr[\mathcal{B} \text{ wins} \mid \lnot\mathsf{E}]
\\&\geq
\Pr[\mathsf{E}] \cdot (1/|S|) + \Pr[\lnot\mathsf{E}] \cdot (\epsilon / |S|).
\end{align*}
So if $\Pr[\mathsf{E}]$ is noticeable, then $\mathcal{B}$ wins with noticeable probability by picking $sk$ (since $|S|$ is poly-sized). And if $\Pr[\lnot\mathsf{E}]$ is noticeable, then $\mathcal{B}$ wins with noticeable probability by outputting $b'$.
Update: This analysis is incomplete. It's not clear why the advantage of $\mathcal{B}$ is noticeable. Here's a failing example.
Consider the following $\mathcal{A}$. Say $\mathcal{A}$ makes two queries $q_0,q_1$ to $H$. In the case $b=1$ $\mathcal{A}$ will make $q_0$ decrypt the ciphertexts to $m_0,m_1$ and $q_1$ decrypt the ciphertexts to $m_1,m_0$. In the case $b=0$, then $\mathcal{A}$ will make both $q_0$ and $q_1$ decrypt the ciphertexts to $m_1,m_0$. Then
\begin{align*}
\Pr[\mathcal{B}(1) = 1] &= \frac{1 + 0 + \Pr[\mathcal{A}(1) = 1]}{3} \\\\
\Pr[\mathcal{B}(0) = 1] &= \frac{1 + 1 + \Pr[\mathcal{A}(0) = 1]}{3} \\\\
\mathsf{Adv}(\mathcal{B}) &= \left|\frac{\Pr[\mathcal{A}(1) = 1] - \Pr[\mathcal{A}(0) = 1] - 1}{3}\right|
\\\\ &\geq |\mathsf{Adv}(\mathcal{A}) - 1/3|
\end{align*}
(the last ineq is reverse triangular inequality).
This is not a lower bound we want, since it vanishes if $\mathcal{A}$'s advantage is $1/3$. The issue might have to do with the supposition that $\mathcal{A}$ can construct that many fake decryption keys, especially ones that decrypt to precisely the opposite thing. It's probably the case that the phrase "the private key" in the question is doing some heavy lifting.
