Let $G$ and $G'$ be groups and suppose $f:G\to G'$ is operation-preserving; i.e. $f(a\ast b) = f(a) \circ f(b)$ where $\ast$ is the group operation on G and $\circ$ is the group operation on $G'$.
The proposition states exactly that $f(e) = e$, but I don't think that this is true (or perhaps, I think it may lack information). It's clear that $e \in G$ and $f(e) \in G'$. However, I don't see why $e_G = e_{G'}$. I can't imagine why the identity element in G must be equal to the identity element in $G'$ if there exists an operation-preserving map between them. I imagine that $f(e_G) = e_{G'}$.
Despite my misunderstanding, I've tried $f(e_G) = f(e_G \ast e_G) = f(e_G) \circ f(e_G)$, and wanted to "add" $-f(e_G)$ to both sides, but I know that's probably incorrect (and also implies $e_G$ is an additive identity element).
Note that: $$ f(e_G) = f(e_G*e_G) = f(e_G)\circ f(e_G) $$ Now, multiply ($G'$ - product) on both sides by the inverse of $f(e_G)$, which exists because $G'$ is a group: $$ e_{G'} =f(e_G) \circ f(e_G)^{-1} = (f(e_G)\circ f(e_G))\circ f(e_G)^{-1} = f(e_G) $$ Which completes the proof.
For the other part, just use the concept of inverse.Note that: $$ e_{G'} = f(e_G) = f(a * a^{-1}) = f(a) \circ f(a^{-1}) $$ So, $f(a)^{-1} = f(a^{-1})$ holds, since their product is the identity.