The result holds in arbitrary vector spaces (even infinite dimension), if we assume the Axiom of Choice for the infinite dimensional case. It will appear in the form of "every linearly independent subset can be extended to a basis".
The key is the following theorem:
Theorem. Let $T\colon V\to W$ be a linear transformation. Then $T$ is one-to-one if and only if for every linearly independent indexed set $\{v_i\}_{i\in I}\subseteq V$, the set $\{T(v_i)\}_{i\in I}$ is linearly independent in $W$.
(I write them as indexed sets to include the case where the set has the same vector multiple times, in which case it is considered to be linearly dependent).
Proof. Assume $T$ is one-to-one, and let $\{v_i\}_{i\in I}$ be linearly independent. Let $i_1,\ldots,i_n$ be distinct indices and let $\alpha_1,\ldots,\alpha_n$ be scalars such that
$$\mathbf{0} = \alpha_1T(v_{i_1}) + \cdots + \alpha_nT(v_{i_n}).$$
Then
$$\begin{align*}
\mathbf{0} &= \alpha_1T(v_{i_1}) + \cdots + \alpha_nT(v_{i_n})\\
&= T(\alpha_1v_{i_1}+\cdots+\alpha_nv_{i_n}).
\end{align*}$$
Since $T$ is one-to-one, it follows that $\alpha_1v_{i_1}+\cdots+\alpha_nv_{i_n} = \mathbf{0}$. Since $\{v_i\}_{i\in I}$ is linearly independent, we conclude that $\alpha_1=\cdots=\alpha_n=0$.
This proves that $\{T(v_i)\}_{i\in I}$ is linearly independent.
Conversely, assume that whenever $\{v_i\}_{i\in I}$ is linearly independent, then $\{T(v_i)\}_{i\in I}$ is linearly independent. We prove that $\ker(T)=\{\mathbf{0}\}$. Indeed, if $v\in\ker(T)$, then $\{T(v)\} = \{\mathbf{0}\}$ is linearly dependent, so it follows that $\{v\}$ is linearly dependent. A one vector set is linearly dependent if and only if $v=\mathbf{0}$, so we conclude that $ker(T)=\mathbf{0}$, as desired. $\Box$
So let $V$ and $W$ be arbitrary vector spaces, let $T\colon V\to W$ be a one-to-one linear transformation.
Let $\beta=\{v_i\}_{i\in I}$ be a basis for $V$. Since $\beta$ is linearly independent and $T$ is one-to-one, we know that $T(\beta)=\{T(v_i)\}_{i\in I}$ is a linearly independent subset of $W$. Extend $T(\beta)$ to a basis $\gamma$ for $W$,
$$\gamma = \{T(v_i)\}_{i\in I}\cup\{w_j\}_{j\in J}.$$
Now define $S\colon W\to V$ by definining it on $\gamma$ and extending linearly. We let $S(T(v_i)) = v_i$ for all $i\in I$, $S(w_j) = \mathbf{0}$ for all $j\in J$.
Then $S\circ T$ and the identity linear transformation $V\to V$ agree on $\beta$, and so they are equal. That is, $S\circ T = \mathrm{I}_V$.
We might note the converse: if $T\colon V\to W$ is a linear transformation, and there exists a linear transformation $S\colon W\to V$ such that $S\circ T=\mathrm{I}_V$, then $T$ is one-to-one.
This is analogous to the set theory result that says that a function $f\colon X\to Y$ with $X\neq\varnothing$ is one-to-one if and only if there exists $g\colon Y\to X$ such that $g\circ f = \mathrm{id}_X$. If you want to get categorical, it is a consequence of the fact that in the category of vector spaces over $k$, every object is free.
