Proposition: Let $R$ be a commutative noetherian ring and let $P$ be a finitely generated $R$-module. Then $P$ is projective if and only if $P_m$ is free for all maximal ideals $m$ of $R$.
Statement: Let $s_1, s_2 \in R$ be such that $Rs_1 +Rs_2 = R$. Let $P_i$ be a finitely generated projective $Rs_i$-module for i = 1, 2. Let $ f $ be an isomorphism from $(P_1){s_2}$ to $(P_2){s_1}$ and let
be a fiber product diagram. Then with help of above proposition, show $P$ is projective $R$ module.
My approach: By definition of fiber product,
$P= \{ (x_1,x_2)\in P_1 \oplus P_2 \colon f(x_1/1)=x_2/1 \}$
Now the maps $\phi_1 \colon P_1 \to P_{s_1}$ and $\psi_1 \colon P_{s_1} \to P_1$ defined as $\phi_1(x_1)=(s_1^i x_1,x_2)/s_1^i$ where $f(x_1/1)=x_2/s_1^i$ and $\psi_1((x_1,x_2)/s_1^i) = \frac{1}{s_1^i} \cdot x_1$ are $R_{s_1}$-module homomorphisms with $\phi_1 \circ \psi_1 = Id_{P_{s_1}}$ and $\psi_1 \circ \phi_1 = Id_{P_1}$ Thus $P_{s_1} \cong P_1$.
Similarly, $P_{s_2} \cong P_2$
So, $P_{s_1}, P_{s_2}$ are finitely generated projective $R_{s_1},R_{s_2}$ modules respectively. Then by definition of projective modules:
$P_{s_i}\oplus Q_i \cong R_{s_i}^{n_i}$, $i=1,2$
Take any maximal ideal $m$ of R. If $s_1,s_2 \in m$ then $Rs_1+Rs_2=R \implies 1\in m$ which is a contradiction. So, both $s_1,s_2$ cannot be in $m$.
Say $s_1 \not\in m$. Then $R_{s_1} \subset R_m$. Then $P_{s_1}\oplus Q_1 \cong R_{s_1}^{n_1} \implies P_{m}\oplus (Q_1\otimes_{R_{s_1}}R_m) \cong R_{m}^{n_1}$ i.e. $P_m$ is projective $R_m$-module.
Assuming $P_1,P_2$ free modules, we get $Q_1,Q_2$ are $0$. Then $P_m$ is free for all $m$.
So P is projective by above Proposition.
My question: how to prove without assuming $P_1,P_2$ are free.
EDIT: I know $Rs_1+Rs_2=R$ implies$ P $is the fiber product of $P_{s_1}$ and $P_{s_2}$ over$ P_{s_1s_2}$. I think it will be of some help.