设函数f(x)=ax-2-lnx(a∈R). (Ⅰ)若f(x)在点(e,f(e))处的切线为x-ey+b=0,求a,b的值; (Ⅱ)求f(x)的单调区间; (Ⅲ)当x>0时,求证:f(x)-ax+ex>0.