I'm not sure I understand the question. More exactly, I think I disagree with what seems to be an assumption built into it: that there are sharp lines to be drawn between 'only using the universal property' and not, or between working 'without elements' and not.
Generally, a universal property describes how something interacts with the world around it. For example, if you say that an object $I$ of a category $\mathcal{C}$ is initial, that describes how $I$ interacts with other objects of $\mathcal{C}$. If you don't know much about the objects of $\mathcal{C}$, it doesn't tell you much about $I$. Similarly, you're not going to be able to deduce anything about the ring $S^{-1}A$ without using facts about rings. I don't know which of those facts are ones you'd be happy to use, and which aren't.
There's nothing uncategorical about elements. For example, you're dealing with rings, and an element of a ring $A$ is simply a homomorphism $\mathbb{Z}[x] \to A$. (And $\mathbb{Z}[x]$ can be characterized as the free ring on one generator, that is, the result of taking the left adjoint to the forgetful functor $\mathbf{Ring} \to \mathbf{Set}$ and applying it to the terminal set 1.)
So I'm unsure what exactly your task is. But I'd like to suggest a different universal property of localization, which might perhaps make your task easier. Here it is.
Let $\mathbf{Set}/\mathbf{Ring}$ be the category of rings equipped with a set-indexed family of elements. Formally, it's a 'comma category'. An object is a triple $(S, i, A)$ where $S$ is a set, $A$ is a ring, and $i$ is a function from $S$ to the underlying set of $A$. (You might think of $i$ as an including $S$ as a subset of $A$, but $i$ doesn't have to be injective.) A map $(S, i, A) \to (S', i', A')$ is a pair $(p, \phi)$ consisting of a function $p: S \to S'$ and a homomorphism $\phi: A \to A'$ making the evident square commute.
There is a functor $R: \mathbf{Ring} \to \mathbf{Set}/\mathbf{Ring}$ given by
$$
R(A) = (A^\times \to A)
$$
where $A^\times$ is the set of units of $A$ and the arrow is the inclusion. Then $R$ has a left adjoint $L$, given by $L(S, i, A) = (iS)^{-1}A$. In other words, the left adjoint to $R$ is localization.
If you want to understand $S^{-1}A$ for any $S\subset A$, you may write $S$ as a filtered union of its finite subsets $S_i$, and it is clear from the universal properties that
$$S^{-1}A=\varinjlim_i \ S_i^{-1}A$$
Therefore it is sufficient to consider the case where $S$ consists of a finite set of elements of $A$. If $f$ denotes the product of all the elements of $S$ in $A$, it is clear that inverting $f$ is the same thing than inverting each element of $S$. Therefore, we may assume that $S$ contains a unique element $f$. It is then obvious (in terms of universal properties) that $S^{-1}A$ is canonically isomorphic to the (filtered) colimit of the diagram indexed by non negative integers
$$A\overset{f}{\to}A\overset{f}{\to}A\overset{f}{\to}\cdots\overset{f}{\to}A\overset{f}{\to}A\overset{f}{\to}\cdots$$
(where $f$ stands for `multiplication by $f$').
Puting back all of the above reductions/descriptions together is the categorical way of writing the usual description of $S^{-1}A$ with elements. (As an exercise, you may turn all this into a global construction, i.e. as one filtered colimit of a diagram which is objectwise just $A$, but in which the transition maps are given by multiplication by a finite product of elements of $S$.)
The problem is now reduced to the understanding of colimits of diagrams of $A$-modules
of the shape
$$M_0\overset{s_1}{\to} M_1 \overset{s_2}{\to} M_2\to\cdots
M_n\overset{s_{n+1}}{\to} M_{n+1}\to\cdots$$
Using the universal properties, one can see that $\varinjlim_n M_n$
is the cokernel of the map
$$1-s:\bigoplus_n M_n\to \bigoplus_n M_n$$
where $s$ sends an element $x$ of $M_n$ to $s_{n+1}(x)$.
In particular, we have a canonical epimorphism
$$\bigoplus_n M_n\to \varinjlim_n M_n$$
This implies that any element of $\varinjlim_n M_n$ comes from an element of $M_n$
for $n$ big enough. This presentation of $\varinjlim_n M_n$
also shows that if an element $x$ of $M_n$ becomes zero in $\varinjlim_n M_n$,
there exists a positive integer $m$ such that $s_{n+m}\ldots s_{n+1}(x)=0$.
Applying this to the diagram
$$A\overset{f}{\to}A\overset{f}{\to}A\overset{f}{\to}\cdots\overset{f}{\to}A\overset{f}{\to}A\overset{f}{\to}\cdots$$
we see that $A[f^{-1}]$ admits the usual description.
N.B. In an abstract context, proving things about localizations consists to use the categorical description above and to use some special properties of filtered colimits (which are often exact, for instance), so that you don't need any description in terms of elements. For instance, the flatness of $S^{-1}A$ comes from the fact that it is a filtered colimit of free $A$-modules (of rank $1$).
One can also compute the kernel of the map $\tau:A\to S^{-1}A$. To keep things simple let us do this in the case where $S$ consists of a single element $f$. Then $ker(\tau)$ is the colimit of $A$-modules
$$\varinjlim_n \; \; ker(A\overset{f^n}{\to}A)$$
(because filtered colimits are exact in the category of $A$-modules).
In conclusion, it seems possible to describe $S^{-1}A$ using categorical arguments for $A$-modules, but I don't see how to obtain such a description using only the theory of rings.
Best Answer
$\newcommand\T{\mathrm T}$I assume that "the universal property" means that, for every $R$-algebra $S$, every $R$-module map $V \to S$ extends uniquely to an $R$-algebra map $\T V \to S$.
In the original version of this answer, I proceeded as follows. For each $v^* \in V^* \mathrel{:=} \operatorname{Hom}_\text{\(R\)-mod}(V, R)$, let $\alpha_{v^*}$ be the unique extension of the map $V \to R[x]$ given by $v \mapsto \langle v^*, v\rangle x$ to an $R$-module map $\T V \to R[x]$. Then, for each $n \in \mathbb Z_{\ge 0}$, $\T^nV$ is the set of $\tau \in \T V$ such that $\alpha_{v^*}(\tau)$ lies in $R x^n$ for all $v^* \in V^*$.
This definition is too coarse: it correctly identifies the degree of each homogeneous element, but, as you pointed out, it can also incorrectly assign elements degrees that they shouldn't have.
The obvious fix is to replace $R$-module maps $V \to R$ by $R$-module maps $V \to A$ for arbitrary $R$-algebras $A$. This works, but the only way that I can see to show that it always works is to take $A$ to be the usual explicitly constructed tensor algebra, or something close to it. For example, I have already cheated a bit by using $R[x]$, which is just the tensor algebra on $R^*$; but I could cheat even more by taking for $A$ the quotient of the polynomial $R$-algebra in $V$-many indeterminates by $r x_v = x_{r v}$. Then there is a natural homomorphism $V \to A$ given by $v \mapsto x_v$, and we could use the above trick on $\T V \to A[x]$, or just observe that $A$ itself is graded (since we're quotienting out by a homogeneous ideal) and use the degree there. But at that point we've practically constructed the tensor algebra.