Let $R$ be a commutative ring. The universal property for the tensor product $A\otimes B$ of two noncommutative $R$-algebras $A,B$ is usually phrased as follows:
"For any $R$-algebra $C$ and for any $R$-algebra morphisms $\varphi_A:A\to C$ and $\varphi_B:B\to C$ satisfying $\varphi_A(a)\varphi_B(b)=\varphi_B(b)\varphi_A(a)$ for all $a\in A$ and $b\in B$, there exists a unique $R$-algebra morphism $\varphi:A\otimes B\to C$ satisfying $\varphi_A=\varphi\circ i_A$ and $\varphi_B=\varphi\circ i_B$."
I don't like the condition $\varphi_A(a)\varphi_B(b)=\varphi_B(b)\varphi_A(a)$ in the definition because it is not arrow-theoretic. Is it possible to rephrase the definition so that it doesn't refer to elements?
As far as I know the answer is basically "no". But if you're happy referring to specific special objects in the category, you can say the following. Given $\varphi_A:A\to C$ and $\varphi_B:B\to C$, their images in $C$ commute iff for any pair of homomorphisms $\psi_A:R[t]\to A$ and $\psi_B:R[t]\to B$, there exists a homomorphism $\psi:R[x,y]\to A$ such that $\psi i_x=\varphi_A\psi_A$ and $\psi i_y=\varphi_B\psi_B$ where $i_x:R[t]\to R[x,y]$ and $i_y:R[t]\to R[x,y]$ are the inclusions sending $t$ to $x$ and $y$, respectively.
(I personally find this approach to be basically cheating, since it just shifts the question to be how you can describe the object $R[x,y]$ together with the inclusions $i_x$ and $i_y$ without elements. But at least it isolates the problem to a single "universal" example.)