Say $R$ is a ring with unit element $1$ and $\phi:R\to R'$ is a ring homomorphism. Prove that $\phi(1)$ is unit element of $R'$.
So I have shown the following with regard to $\phi(1)$
$$\phi(1)\cdot\phi(a)=\phi(a)\cdot\phi(1)=\phi(a)$$ $$ \phi(-1)=-\phi(1)$$
Is this sufficient, or there is something more I need to prove?
If $\phi: R \to R'$ is surjective, then there exists some $x \in R$ such that $\phi(x) = 1_{R'}$. Hence, observe that $$\phi(1_R)\cdot \phi(x) = \phi(x) = 1_{R'}$$ and obviously $$ \phi(x)\cdot \phi(1_R)= \phi(x) = 1_{R'}$$ where homomorphism properties were applied. Note both ways were checked, since such a check is required to verify that an element is a unit.
Therefore, $\phi(x) = 1_{R'}$ is the multiplicative inverse of $\phi(1_R)$ which makes it a unit. Note more generally that this proves that $\phi(1_R) = 1_{R'}$, since $\phi(1_R) = \phi(1_R)\cdot 1_{R'} = \phi(1_R)\cdot \phi(x) = 1_{R'}$.
Just for your information, this also works if $R'$ is an integral domain. In fact, if $\phi(1_R)\not= 1_{R'}$, then $\phi(1_R)$ is actually a zero divisor of $R'$ or $\phi$ is the zero map! Observe that $$ \phi(1_R^2) - \phi(1_R) = 0 \implies \phi(1_R)\phi(1_R) - \phi(1_R) = 0 \implies (\phi(1_R) - 1_{R'})\phi(1_R) = 0. $$ If you suppose that $\phi(1_R) \not= 1_{R'}$, the only solution is if $\phi(1_R)$ is a zero divisor (it turns out $\phi(1_R) \not= 0$ unless $\phi$ is the zero map).
Suppose however that $R'$ has no zero divisors; i.e. it's an integral domain! Then we have to have that $$ \phi(1_R) = 1_{R'} $$ and hence $\phi(1_R)$ is a unit. So this also works if $R'$ is an integral domain.