The answer is yes.
Indeed, let us write $X$ instead of $x$, according to standard notation, to distinguish between random variables (denoted by upper-case letters) and their values (denoted by lower-case letters). Let us write $P$ instead of $\Pr$, and then let us also write $A$ instead of $E$, to distinguish it from the expectation sign.

Then we need to show that
\begin{equation}
EX\,P(A|X)\ge P(A)\,EX
\end{equation}
given that
\begin{equation}
P(A|X>y)\ge P(A)
\end{equation}
for all $y$ such that $P(X>y)\ne0$.

Replacing $X$ by $X-a$, we may assume that $X\ge0$. Then
\begin{multline*}
EX\,P(A|X)=EX\,E(1_A|X)=EX\,1_A=E\Big(\int_0^X dy\Big)\,1_A \\
=E\Big(\int_0^\infty dy\,1_{X>y}\Big)\,1_A
=E\Big(\int_0^\infty dy\,1_{X>y}1_A\Big) \\
=E\int_0^\infty dy\,1_{X>y,A}=\int_0^\infty dy\,E1_{X>y,A} \\
=\int_0^\infty dy\,P(X>y,A) =\int_0^\infty dy\,P(A|X>y)P(X>y) \\
\ge\int_0^\infty dy\,P(A)P(X>y)=P(A)\int_0^\infty dy\,P(X>y)=P(A)EX,
\end{multline*}
as claimed.

(The last equality follows immediately from the special case $EX=\int_0^\infty dy\,P(X>y)$ of the previously proved equality $EX\,P(A|X)=\int_0^\infty dy\,P(X>y,A)$, with $A$ replaced there by an event of probability $1$.)