Cell notation for the lead-acid battery
There are a couple of things wrong here. First off, your final reaction is unbalanced. Once you've fixed the balancing, read the other mistakes:


The ions do not exist in the liquid state! They are solvated/hydrated by the solvent. Since the solvent is water here, we'll say that the ions are in the aqueous (aq) phase instead. 
While there is certainly no salt bridge here, there is still an electrolyte - an aqueous solution of sulphuric acid. Hence, you must mention it in your cell notation, between the anode and the cathode.
The $\ce{PbSO4}$ formed at the anode is in solid state. Hence, writing it as $\ce{Pb^2+(s)}$ is incorrect, as it is not dissociated into the ions $\ce{Pb^2+}$ and $\ce{SO4^2-}$.


With all these corrections, your final, correct cell representation should be:


   $$\small{\ce{Pb(s), HSO4^-(aq) | PbSO4(s), H+ | H2SO4 ($\pu{x~M}$) | PbO2(s), HSO4-(aq), H+(aq) | PbSO4(s)}}$$




Some websites (like KhanAcademy) and texts (NCERT 12), cite the lead-acid battery reaction as this instead:

$$
\begin{array}{}
\text{Oxidation}&amp;\ce {Pb(s) + SO4^2-(aq) &amp;-&gt; PbSO4(s) + 2e-}\\
\text{Reduction}&amp;\ce{PbO2 + SO4^2-(aq) + 4H+(aq) + 2e- &amp;-&gt; PbSO4(s) + 2H2O(l)}\\
\text{Total reaction}&amp;\ce{Pb(s) + PbO2(s) +2SO4^2(aq)- + 4H+(aq) &amp;-&gt; 2PbSO4(s) + 2H2O(l)}\\
\end{array}
$$ 

assuming the bisulphite ion to be further ionized. The cell notation in this case would then be:


   $$\small{\ce{Pb(s), SO4^2-(aq) | PbSO4(s) | H2SO4 ($\pu{x~M}$) | PbO2(s), SO4^2-(aq), H+(aq) | PbSO4(s)}}$$