Suppose $G$ is a group, $R$ is the set of elements of a row of $G$'s cayley diagram, $g, h$ are elements of $G$, and $r$ is an element of $R$ that satisfies $gh = r$. Show $R$ contains every element of $G$ only once.

$Proof$ By $gh = r$ there exists an injection $f_1: G \to R$. $gh = r$ implies by group properties $h = g^{-1}r$, so there is a surjection $f_2: G \to R$.

any way to get a bijection to end this proof, without further defining $f_1, f_2$?

I would start with:
Suppose G is a group, R is the set of elements of a row of G's cayley diagram. Show R contains every element of G only once.

For a fixed element $g \in G$, the row is the map $h \mapsto g \cdot h$. Let me call the map $f$ as in $f(h) = g \cdot h$.
you need to show that if $f(h_1) = f(h_2)$ then $h_1 = h_2$.

First show bijectivity, and then:
Let g be an element of a group G. By definition of a group, g is invertible, so group multiplication by g is bijective. An element r of the row R of a cayley table is an image of group multiplication by g. Therefore, by bijectivity, an element of G appears in R exactly once. ■

Let $G$ be a group with $\cdot$ and $R$ be the set of elements in a row of $G$'s cayley table. For $g, h \in G$ and $r \in R$, we know $g\cdot h = r$. Show that $R$ contains each element of $G$ exactly once.
**Proof.** Since group multiplication is a map $f:G \to G$ that assigns $g,h$ to a unique $g\cdot h$ in $G$, and since $g\cdot h = r$, $r$ is identical to a unique $g\cdot h$ in $G$. Since $r$ is any element in $R$, $R$ contains each element of $G$ exactly once.