Is it possible to prove that $\text{tr}(AB)=\text{tr}(BA)$ without using matrices or having to choose a particular base ?
Such a proof should probably use a non matricial definition of traces. One could say for example that $\text{tr}(A)$ is the linear coefficient of the characteristic polynomial $\det(A-X\text{Id})$. In case the latter doesn't satisfy everyone, here is a definition of trace suggested in Halmos's Finite Dimensional Vector Spaces. Let $E$ be a $n$-dimensional vector space over some field $k$. We recall that the space $W$ of alternating $n$-linear forms $\omega: E^{n}\rightarrow k$ is one dimensional over $k$. Given an endomorphism $f$ of $E$ we can define the linear application \begin{align*} \bar{f}:\quad W &\rightarrow W \\ \omega(x_{1},x_{2},...) &\mapsto \sum_{i} \omega(x_{1},...,x_{i-1},f(x_{i}),x_{i+1},..) \end{align*} Since $W$ is one dimensional $\bar{f}=\lambda Id$. We set $\text{tr}(f)=\lambda$.
The analog definition of the determinant easily yields $\det(AB)=\det(A)\det(B)=\det(BA)$, but this no longer works here. One could fix a base $e_{1},e_{2},..,e_{n}$ and do the calculations but this would kind of beat the point, since it is essentially the same as computing with matrices.
