My attempt: $P$ is projective. Take $0 \to B \to I_0 \to I_1 \to 0$ is an injective resolution of $B$. Apply $Hom(P,\_)$, left exactness of covariant functor implies exactness of the sequence: $0 \to Hom(P,B) \to Hom(P,I_0) \to Hom(P,I_1)$. Note $Hom(P,I_0) \to Hom(P,I_1)$ is onto. Then $Ext^1(P,B) = \frac{ker(Hom(P,I_1) \to Hom(P,I_2))}{im(Hom(P,I_0) \to Hom(P,I_1))}=\frac{Hom(P,I_1)}{Hom(P,I_1)}=0$ because $I_2$ doesn't exist so $ker(Hom(P,I_1) \to 0)=Hom(P,I_1)$.
Is this right in any way? I did see solutions that use $P$ as a direct summand but I just wanted to use the definition of $Ext(\_,\_)$. Bredon just straight up says that because $Hom(P,I_0) \to Hom(P,I_1)$ is onto, then $Ext(P,B)=0$ without splitting argument (or is it implied?) and I didn't understand how.
This does not quite work. First, note that $B$ may not have an injective resolution of the form $0\rightarrow B\rightarrow I_0\rightarrow I_1\rightarrow 0$, as this would imply that certain Ext modules are trivial, since Ext will be the homology of the sequence below: $$0\rightarrow \text{Hom}(A,I_0)\rightarrow\text{Hom}(A,I_1)\rightarrow 0\rightarrow \dots$$ In particular, $\text{Ext}^n (A,B)$ will be trivial for all $n \geq 2$ for any module $A$. But there are modules with nontrivial higher Ext modules, so the injective resolution you wrote above does not always work.
EDIT (Explanation written out more explicitly)
Note that if $B$ has a resolution of the desired form, the r.h.s. zero extends indefinitely, i.e., the resolution looks like $0\rightarrow B\rightarrow I_0\rightarrow I_1\rightarrow 0\rightarrow0\rightarrow\dots$, but oftentimes we truncate it after the first zero. Since $0$ is a zero object, $\text{Hom}(A,0)=0$ for any module $A$, so the Ext modules are the homology of the sequence: $$0\rightarrow\text{Hom}(P,I_0)\rightarrow\text{Hom}(P,I_1)\rightarrow0\rightarrow0\rightarrow\dots$$ Assuming $P$ to be projective, note that $\phi:I_0\rightarrow I_1$ is surjective (this is just the map given in the resolution of $B$), and since $P$ is projective, given any morphism in $\psi\in\text{Hom}(P,I_1)$, there exists some morphism in $\tilde{\psi}\in\text{Hom}(P,I_0)$ such that $\phi\circ\tilde{\psi}=\phi_*(\tilde{\psi})=\psi$, i.e., $\phi_*:\text{Hom}(P,I_0)\rightarrow\text{Hom}(P,I_1)$ is surjective, so $\text{im}\hspace{.5mm}\phi_*=\text{Hom}(P,I_1)$ . Since the next term in the sequence is the zero module (which is a terminal object), we know that the kernel of $\text{Hom}(P,I_1)\rightarrow 0$ must be $\text{Hom}(P,I_1)$, so $\text{Ext}^1(P,B)=\text{Hom}(P,I_1)\hspace{.5mm}/\hspace{.5mm}\text{Hom}(P,I_1)=0$.
So, $I_2$ still very much exists: it is the zero module. Apart from this point, your argument is correct. The biggest thing to remember is that if a resolution truncates with $0$, we can extend it with infinitely more $0$ modules since $0\rightarrow 0\rightarrow 0$ is always exact ($0$ is both an initial and terminal object).
I would go about doing this in full generality by taking a projective resolution of a projective $P$ of the form: $$\dots\rightarrow 0\rightarrow P\rightarrow P\rightarrow 0$$ And applying $\text{Hom}(-,B)$, we get that the Ext modules are the homology of: $$0\rightarrow \text{Hom}(P,B)\rightarrow 0\rightarrow\dots$$ From which it is clear that $\text{Ext}^1(P,B)=0$ for all modules $B$. Using the fact that $\text{Ext}^*(P,-)$ is a homological $\delta$-functor, it is straightforward to show that all the higher Ext modules are trivial.
Hope this helps!