Finite rings of prime order must have a multiplicative identity

Let $x$ be a nonzero element of the ring. Then $R=\{0,x,2x,3x,\ldots,(p-1)x\}$ where $2x$ means $x+x$ etc. Then $x^2=jx$ where $1\le j\le p-1$. Moreover $(ax)(bx)=abx^2=(abk)x$. All you need to do is to prove that for some $a$, $(abk)x=bx$ for all $b$. (It's surely enough to do this for $b=1$).