Yes, it is really necessary that $P\neq Q$.
Indeed, if $V=\mathbb A^1_k\setminus \lbrace P\rbrace$ , then $H^1(\mathbb A^1_k,\mathbb Z_V) =0$ because the sheaf $\mathbb Z_V$ on $\mathbb A^1_k$ is flasque and flasque sheaves are acyclic.
This follows from the explicit description of its groups of sections, which is easy if you remember that $\mathbb Z_V \subset \mathbb Z$:
The sections of $\mathbb Z_V$
-For an open subset $W\subset \mathbb A^1_k$ not containing $P$, that is $W\subset V$, we have $\Gamma (W,\mathbb Z_V)=\mathbb Z$ , since $\mathbb Z_V|V=\mathbb Z $ as sheaves on V .
-For an open subset $W\subset \mathbb A^1_k$ containing $P$ we have $\Gamma (W,\mathbb Z_V)=\mathbb 0$, since $\Gamma (\mathbb A^1_k,\mathbb Z_V)\subset \Gamma (\mathbb A^1_k,\mathbb Z)$ and the stalk of $\mathbb Z_V$ at $P$ is zero: $(\mathbb Z_V)_P=0$
A variant One could also use the long exact sequence associated to $0\to \mathbb Z_V \to \mathbb Z \to j_\ast (\mathbb Z|\lbrace P\rbrace) \to 0 \quad ( \ast)$ and get: \quad 0\to \text {don't care}\to \Gamma (\mathbb A^1_k,\mathbb Z)=\mathbb Z \stackrel {=}{\to}\Gamma (\mathbb A^1_k,j_\ast (\mathbb Z|\lbrace P\rbrace) =\mathbb Z\to H^1(\mathbb A^1_k,\mathbb Z_V)\to H^1(\mathbb A^1_k,\mathbb Z)=0 \to \cdots The notable points are that $H^1(\mathbb A^1_k,\mathbb Z)=0$ because constant sheaves are flasque on irreducible spaces and above all that the morphism $\Gamma (\mathbb A^1_k,\mathbb Z)=\mathbb Z \to \Gamma (\mathbb A^1_k,j_\ast (\mathbb Z|\lbrace P\rbrace) =\mathbb Z$ is equality, which results from the definition of the maps in $(\ast)$.
(I think this variant is what you wanted to write in your analysis, but your first "$\mathbb Z$" [where I wrote "don't care" !] should actually be zero, and you shouldn't tensor by $\mathbb Q$)