I think that the answer is **no**, already for ruled surfaces over an elliptic curve.

In fact, let $C$ be a curve of genus $\geq 1$ and let us consider two distinct points $x, \, y \in C$. Take now the two rank $2$ vector bundles on $C$ given by $$\mathscr{E}_x = \mathscr{O}_C \oplus \mathscr{O}_C(x), \quad  \mathscr{E}_y = \mathscr{O}_C \oplus \mathscr{O}_C(y).$$

I claim that $\mathbb{P}(\mathscr{E}_x)$ and $\mathbb{P}(\mathscr{E}_y)$ are *not* isomorphic. 

By contradiction, assume that they are isomorphic. Then there exist a line bundle $\mathscr{L}$ on $C$ such that $\mathscr{E}_x \otimes \mathscr{L}=\mathscr{E}_y$, that is $$\mathscr{L} \oplus \mathscr{L}(x) =  \mathscr{O}_C \oplus \mathscr{O}_C(y).$$

Since by the Krull-Schmidt theorem the decomposition of a vector bundle into indecomposable summands is unique up to permutations of the summands, we must have either $\mathscr{L} = \mathscr{O}_C$ or $\mathscr{L}=\mathscr{O}_C(y).$    

Bot cases are impossible: the former since since $x \neq y$ and the curve has positive genus, and the latter because it implies $\mathscr{O}_C(y) \oplus \mathscr{O}_C(x+y)= \mathscr{O}_C \oplus \mathscr{O}_C(y).$