Assume the surjective ring homomorphism $\varphi: R \longrightarrow R'$, where $R$ is a ring with unity. Also let $u$ be an invertible element of $R$. Show that $\varphi(u)$ is invertible in $R'$ iff no invertible element of $R$ belongs to $\ker \varphi$.
$\Rightarrow:$ Suppose $\varphi(u)$ is invertible in $R'$. Since $0$ cannot be invertible (unless $R = \{0\}$, but then we wouldn't have $1$), we have that $$ \varphi(u) \neq 0 \Rightarrow u \notin \ker \varphi $$
$\Leftarrow:$ Since $\varphi$ is surjective, for every $r' \in R'$, there exists $r \in R$ such that $\varphi(r) = r'$ and thus $$ r'\varphi(1) = \varphi(r)\varphi(1) = \varphi(r\cdot 1) = \varphi(r) = r' \Rightarrow \varphi(1) \,\text{unity of}\, R' $$
Also, since $u$ is invertible in $R$, there exists an $r \in R$ such that $ur = 1$ and since $\varphi(u) \neq 0$
$$ \varphi(u)\varphi(r) = \varphi(ur) = \varphi(1) $$ Therefore $\varphi(r)$ is the inverse of $\varphi(u)$ and thus $\varphi(u)$ is invertible.
If $R$ is a unitary ring (i.e. with 1), denote by $R^*$ the set of its invertible elements, if the ring has no unity then set $R^*=\emptyset$.
In your statement you need to assume that both $R$ and $R'$ are rings with $1$ and that $\varphi(1_R)=1_{R'}$ (i.e. $\varphi$ is an homomorphism of unitary rings), otherwise $\varphi : \mathbb{Z} \rightarrow 2\mathbb{Z}$ defined as $\varphi(x)=2x$ gives you a counterexample as $\varphi$ satisfies the conditions but $\varphi(1)=2$.
Now about $\Rightarrow$
note that $\ker(\varphi)$ is an ideal, hence if it contains an invertible element then it is $\ker(\varphi)=R$ and $\varphi=0$. So if you assume $\varphi \neq 0$ then it has to be $\ker(\varphi) \cap R^*=\emptyset$.
if you still don't know what ideals are, then to conclude the same proceed like this: if there were $v \in R^*$ s.t. $\varphi (v) = 0 $ then given any $x \in R$ you would have $\varphi(x)=\varphi(x v v^{-1})= 0$, thus one would have $\varphi=0$.
The only way you could have the condition $\ker(\varphi) \cap R^*=\emptyset$ to fail is that $\varphi=0$. Whether this is possible for a unitary ring homomorphism much depends on what you admit to be a unitary ring. You may want to regard $\{0\}$ as a unitary ring in which $1=0$: it depends on whether you put the axiom $1 \neq 0$ in the theory of unitary rings. If you admit $\{0\}$ as "degenerate" ring with unity, then $0: R \rightarrow \{0\}$ is a unitary ring homomorphism, otherwise all homomorphism of unitary rings $\varphi: R \rightarrow R'$ are $\varphi \neq 0$.
About $\Leftarrow$