I think under some assumption (your conditions might already suffice) you may use excision. 

Pick an open set $U$ on $X$ such that the vector bundle $E$ is trivial, denote $X-U=:Z$, and write $Z$ to be $Z_{1}$, which is a divisor, and some higher codimensional stuff, say $Z_{\geq 2}$. Then using excision, higher codimensional terms may be ignored, namely we may assume $Z_{\geq 2}=0$. Also by induction we may assume $Z=Z_{1}$ is a prime divisor. Then we apply excision to $(X,Z)$ get two exact sequences:
$$\mathbb{Z} \rightarrow \mbox{Pic}(X) \rightarrow \mbox{Pic}(X-Z)\rightarrow 0$$
and
$$\mathbb{Z} \rightarrow \mbox{Pic}(\mathbb{P}(E))\rightarrow \mbox{Pic}(\mathbb{P}(E|_{X-Z}))\rightarrow 0$$
Now we can modify the first sequence by adding $\mathbb{Z}$ terms to the second and third term  with the identity map between them, then the sequence is still exact, namely we have
$$\mathbb{Z} \rightarrow \mbox{Pic}(X)\oplus\mathbb{Z} \rightarrow \mbox{Pic}(X-Z)\oplus\mathbb{Z}\rightarrow 0$$ 
And for the second sequence, since $E|_{X-Z}$ is trivial, by an exercise on Hartshorne, $\mbox{Pic}(Y\times \mathbb{P}^{n})=\mbox{Pic}(Y)\oplus \mathbb{Z}$, apply this to the second SES we have
$$\mathbb{Z} \rightarrow \mbox{Pic}(\mathbb{P}(E))\rightarrow \mbox{Pic}(X-Z)\oplus\mathbb{Z}\rightarrow 0$$
Now the first and third terms of these two sequences are equal, and there is a map on the second position, denoted, say, $\pi^{*}\oplus \phi$, where $\pi^{*}$ is the pull back of $\pi: \mathbb{P}(E)\rightarrow X$, and $\phi$ sends $1$ to $\mathcal{O}(1)$.

Now then the result follows from five lemma if we can show $\pi^{*}$ is injective, since $\phi$ is already injective. By abusing of notation we denote the total space of vector bundle $E$ by $E$ and $i: X\rightarrow E$ be the inclusion, we get pullback $$i^{*}: \mbox{Pic}(E)\rightarrow \mbox{Pic}(X).$$ Also, the projection $p: E-X \rightarrow \mathbb{P}(E)$ gives us another pullback $$p^{*}: \mbox{Pic}(\mathbb{P}(E))\rightarrow \mbox{Pic}(E-X).$$ But we may of course assume rank of $E$ is $\geq 2$, then $X$ has codimension $\geq 2$ in $E$, which implies $$\mbox{Pic}(E-X)=\mbox{Pic}(E).$$ These maps concatenate to a map $\mbox{Pic}(\mathbb{P}(E))\rightarrow \mbox{Pic}(X)$, which is easily seen to be a section of $\pi^{*}$. Hence $\pi^{*}$ is injective.