Let $T,S$ be monoids. A partial surjective mapping $\psi : T \to S$ is called a cover map if for each $s \in S$ there exists some $\hat{s} \in T$ called a cover of $s$ such that for each $t \in \mbox{dom}(\psi)$ we have $$ \psi(t) \cdot s = \psi(t \cdot \hat{s}). $$ Now if there exists a cover map between $T$ and $S$, then there exists a homomorphism of a submonoid of $T$ onto $S$. For the proof set $$ U := \{ \hat{s} \in T : s \in S \} $$ the set of all covers of elements from $S$. This set is a submonoid for if $\hat{a}, \hat{b} \in U$ and $t \in \mbox{dom}(\psi)$ then $\psi(t) ab = \psi(t \hat{a}) b = \psi((t \hat{a}) \hat{b}) = \psi(t (\hat{a}\hat{b})$ and therefore $\hat{a}\hat{b}$ is a cover of $ab$. Furthermore $\psi(t) \cdot 1 = \psi(t \cdot 1)$ and therefore the identity in $T$ is a cover of the identity in $S$, i.e. $1_T = \hat{1_S}$. Now define the map $\varphi : U \to S$ by $\varphi(\hat{s}) = s$. Because each $s \in S$ has a cover this map is surjective, now we have to show that it is well-defined. For this suppose $\hat{s} = \hat{s'}$ and choose by surjectivity of $\psi$ some $n \in T$ with $\psi(n) = 1_S$. Then $$s = 1_S \cdot s = \psi(n) \cdot s = \psi(n \cdot \hat{s}) = \psi(n \cdot \hat{s'}) = \psi(n) \cdot s' = 1_S \cdot s' = s'$$ and so $s = s'$.
Now I want to know if this result also holds for semigroups $T,S$?
I don't think this will work as-is for semigroups, because without access to an identity element in $S$ you will not be able to isolate $s$ in the equation $\psi(t)\cdot s=\psi(t\cdot\hat s)$ to show that $\varphi$ is well-defined. There could of course be some other definition of $U$ and $\varphi$ that works, but it seems unlikely to me.
However, it looks like you need to only change it a little bit to make it work for semigroups: assume that $\psi\colon T\to S^1$ is surjective, rather than $\psi\colon T\to S$, then you can find your $n\in T$ with $\psi(n)=1_S$ and everything should work as before. Note that $S^1$ denotes the monoid formed by adjoining an artificial identity element $1_S$ to $S$.
Slightly off-topic: If you want to avoid picking each $\hat s$ arbitrarily (it might not really be arbitrarily in your particular context) when you form $U$ then you could define a submonoid of $T$ in the following way. For each $s\in S$ take $C(s)\subseteq T$ to be the set $$ C(s)=\{u\in T:\text{$\psi(t)\cdot s=\psi(t\cdot u)$ for all $t\in\operatorname{dom}(\psi)$}\}. $$ That is, $C(s)$ is the set of all covers of $s$, and so it clearly contains your distinguished cover $\hat s$. You could then consider the submonoid $$ \bigcup_{s\in S}C(s) $$ of $T$. Notice that this is a submonoid because (as you essentially showed) if $u_1\in C(s_1)$ and $u_2\in C(s_2)$ then $u_1\cdot u_2\in C(s_1\cdot s_2)$.
In a way this submonoid captures all the "covering behaviour" of $\psi$, not just a particular chosen "slice" of it. But in the end not a great deal changes; I thought you might find the perspective interesting though.