Is there a quotient of a symmetric algebra which is in bijection with the simple tensors? Does there exist a nonzero vector space $V$ over a field $F$ and a quotient $\overline{S(V)}$ of the symmetric algebra on $V$ such that the quotient map $\mathfrak{S} \to \overline{S(V)}$, where $\mathfrak{S} \subseteq S(V)$ denotes the set of simple tensors, is a bijection?
In other words, the elements of $\overline{S(V)}$ would be represented, and uniquely so, by simple tensors. This makes $\overline{S(V)}$ into a UFD with primes $V - \{0\}$ and units $F^\times$. Conversely, given a UFD $R$ with nonempty set of primes $P$ such that $P ~ \cup ~ \{0\}$ is a group under addition, we find that $V := P ~ \cup ~ \{0\}$ is in fact a vector space over the field $F := R^\times ~ \cup ~ \{0\}$, and $R$ gives rise to such a quotient $\overline{S(V)}$ as above. So the question can be rephrased: "Does there exist a non-field UFD whose set of primes together with $0$ form an additive group?"
Beyond this equivalence, there's not much that's useful that I can say about this problem. Does such a ring/algebra quotient exist?
 A: Yes, such a quotient does exist.  Let me start with a lemma.

Lemma: Let $A$ be a UFD, let $a\in A$ be a prime, and let $B=A[x,y,z]/(xyz-a)$.  Then:

*

*$B$ is a UFD.


*$B^\times=A^\times$ (i.e., every unit of $B$ was already a unit of $A$).


*Let $b\in B$ be a nonzero element of the form $ux+vy+wz+c$ where $u,v,w\in A^\times\cup\{0\}$ and $c\in A$ is either $0$ or prime and not associate to $a$.  Then $b$ is prime in $B$.

Proof: Since $A[x,y,z]$ is a UFD and $xyz-a$ is irreducible in $A[x,y,z]$, $B$ is a domain.  Note also that $B/(x)\cong A[y,z]/(a)\cong (A/a)[y,z]$ is a domain, so $x$ is prime in $B$, and by symmetry so are $y$ and $z$.
Note that $B$ is free as an $A$-module with basis consisting of all monomials in $x,y,$ and $z$ which do not contain $xyz$.  We can consider $B$ as a $\mathbb{Z}^2$-graded ring by taking $A$ to have degree $(0,0)$, $x$ to have degree $(1,0)$, $y$ to have degree $(0,1)$, and $z$ to have degree $(-1,-1)$.  Each of our monomial basis elements then has a distinct degree, so a homogeneous element is a constant multiple of a monomial (where "constant" means element of $A$).  Given an element $b\in B$, we define the support $\operatorname{supp}(b)$ as the convex hull in $\mathbb{R}^2$ of the set of degrees in which $b$ has a nonzero homogenous part.  The support is a (possibly degenerate) convex polygon whose vertices have integer coordinates.  Note that if $b,c\in B$ then $$\operatorname{supp}(bc)=\operatorname{supp}(b)+\operatorname{supp}(c)$$ (here the sum on the right is the operation $S+T=\{s+t:s\in S,t\in T\}$ on subsets of $\mathbb{R}^2$).  Indeed, note that $\operatorname{supp}(b)+\operatorname{supp}(c)$ is a convex polygon, and each of its vertices is an extreme point and thus is a degree that can only be obtained in one way by multiplying terms of $b$ and terms of $c$.  This means there can be no cancellation, so every vertex of $\operatorname{supp}(b)+\operatorname{supp}(c)$ really is in the support of $bc$.
Using this property of supports, we can immediately conclude that $B^\times=A^\times$ (since a unit of $B$ must have singleton support).  We can also see that any element $b$ as in statement (3) is irreducible, since its support cannot be written nontrivially as a sum of two convex polygons with integer vertices.  This means we need only consider factorizations where one factor is homogeneous, and it is easy to check that no nontrivial such factorization can exist.
[You may be wondering why this Lemma defines $B=A[x,y,z]/(xyz-a)$, instead of something like $B=A[x,y]/(xy-a)$ or more generally $B=A[x_1,\dots,x_n]/(x_1\dots x_n-a)$ for some other $n$.  The reason is the step above: if we used only two variables, then statement (3) might fail, since the shapes of the supports of such elements $b$ would be different and would allow them to be factored.  For instance, if $a+1$ happens to be prime, then $x+y+(a+1)$ could be factored as $(x+1)(y+1)$.  The argument would still work if you used more than $3$ variables, though.]
We can also use supports to see that $B$ satisfies the ascending chain condition on principal ideals (proof sketch: in a chain of principal ideals, the support can only shrink finitely many times, and so eventually we are just factoring out homogeneous elements each time, so we can reduce to the fact that $A$ satisfies the ascending chain condition on principal ideals).
To complete the proof of the Lemma, all that remains is to show that every irreducible element of $B$ is prime.  We have already shown this for $x,y,$ and $z$, so suppose $b\in B$ is an irreducible element that is not associate to $x,y,$ or $z$.  Then in particular, $b$ is not divisible by $a$, so one of its homogeneous parts is not divisible by $a$.  We assume without loss of generality that $z$ does not appear in this homogeneous part.
Now consider $B$ as a subring of the UFD $C=A[x,y,x^{-1},y^{-1}]$ by mapping $z$ to $\frac{a}{xy}$.  For any $c\in C$, we have $a^nc\in B$ for some $n\in\mathbb{N}$.  Also, $b$ is still not divisible by $a$ in $C$, since $z$ does not appear in its term which is not divisible by $a$.  I now claim $b$ is irreducible in $C$.  Indeed, suppose $b=cd$ for $c,d\in C$ which are not units.  We can then write $a^{n+m}b=(a^nc)(a^md)$ where $a^nc,a^md\in B$.  But $a^{n+m}=(xyz)^{n+m}$ is a product of prime elements of $B$, so $a^nc$ and $a^md$ must collectively have $n+m$ factors of each of $x,y,$ and $z$ in $B$.  We can thus divide out these factors and get a factorization of $b$ in $B$, which must be trivial since $b$ is irreducible in $B$.  This means that either $a^nc$ or $a^md$ is just a product of copies of $x,y,$ and $z$ (up to units), which means it is just a power of $a$ times a unit of $C$.  But $b$ is not divisible by $a$ in $C$, so this means $c$ or $d$ must be just a unit of $C$, which is a contradiction.
So, $b$ is irreducible (and thus prime) in $C$.  Now suppose $b$ divides $cd$ for some $c,d\in B$.  Since $b$ is prime in $C$, it divides either $c$ or $d$ in $C$; let us say it divides $c$.  We then can write $c=be$ for some $e\in C$ and so $a^nc=b(a^ne)$ where $a^ne\in B$.  But now $a^n=(xyz)^n$ is a product of primes of $B$ that do not divide $b$, so it must divide $a^ne$ in $B$.  We conclude that actually $e\in B$ and so $b$ divides $c$ in $B$.  This completes the proof that $b$ is prime in $B$, which completes the proof of the Lemma.

Now we construct an example of an algebra with your property.  The idea is to iterate the construction of the Lemma in order to eliminate all primes that are not simple tensors, while using part (3) of the Lemma to guarantee that all the degree $1$ simple tensors remain prime.  Let $F$ be a field and $X_0$ be a nonempty set.  Let $V_0$ be a vector space with $X_0$ as a basis and $A_0=S(X_0)$.  Let $P_0$ be a set which contains one representative of each associate class of prime element of $A_0$ which is not a simple tensor.  We now define $$A_1=A_0[x_p,y_p,z_p]/(x_py_pz_p-p)$$ where $p$ ranges over all elements of $P_0$ (so we're adjoining separate elements $x_p,y_p,$ and $z_p$ for each $p\in P_0$).  We consider $A_1$ as a quotient of $S(V_1)$, where $V_1$ has as a basis $X_0\cup\{x_p,y_p,z_p:p\in P_0\}$.
For any finite subset $Q\subseteq P$, let $A_0(Q)$ be the $A_0$-subalgebra of $A_1$ generated by the $x_p,y_p,$ and $z_p$ for $p\in Q$.  Note that such a ring $A_0(Q)$ can be built from $A_0$ by iterating the construction of the Lemma finitely many times, once for each element of $Q$.  It follows that $A_0(Q)$ is a UFD and its units are just $F^\times$.  Part (3) of the Lemma also tells us that every element of $V_1$ which is contained in $A_0(Q)$ is prime in $Q$.  Moreover, note that if $Q'\supseteq Q$, then every prime of $A_0(Q)$ remains a prime of $A_0(Q')$ except for the elements of $Q'\setminus Q$ which each become a product of three new primes.  It follows that any prime of $A_0$ which is a simple tensor remains irreducible in $A_1$, as does any prime of $A_0(Q)$ which was not in $A_0$.  We can now see that any element $b\in A_1$ has a factorization into irreducible elements: there exists $Q$ such that $b\in A_0(Q)$, and we can factor $b$ in $A_0(Q)$.  Some of its factors may be non-simple tensors of $A_0$, but we can pass to a $Q'\supset Q$ such that all of them factor in $A_0(Q')$, and then all the irreducible factors of $b$ in $A_0(Q')$ remain irreducible in $A_1$.  Moreover, this factorization is unique, since if we had two different such factorizations, we could find a $Q''$ such that both of them are contained in $A_0(Q'')$, contradicting the fact that $A_0(Q'')$ is a UFD.
The upshot is that $A_1$ is a UFD, its group of units is just $F^\times$, and every element of $V_1$ is prime in $A_1$.  Moreover, all the irreducible elements of $A_0$ which were not simple tensors now factor into simple tensors in $A_1$.
We can now repeat this construction starting with $A_1$ instead of $A_0$, to get an extension $A_2$ of $A_1$ with the same properties.  Namely, $A_2$ which is a UFD with unit group $F^\times$.  It is a quotient of a symmetric algebra $S(V_2)$ for some $V_2\supseteq V_1$ such that every element of $V_2$ is prime in $A_2$, and all the irreducible elements of $A_1$ which were not simple tensors now factor into simple tensors in $A_2$.
We repeat this construction to recursively build a chain of UFDs $$A_0\subseteq A_1\subseteq A_2\subseteq A_3\subseteq\dots$$ in the same way.  Now let $A$ be the direct limit of this chain.  Then $A^\times=F^\times$, and $A$ is a UFD, since the prime factorization of any element of $A_n$ stabiliizes after $A_{n+1}$ (since its factorization in $A_{n+1}$ uses only simple tensors, which will remain prime in each later stage of the construction).  We can consider it as a quotient of a symmetric algebra $S(V)$ where $V$ is the direct limt of the vector spaces $V_n$ corresponding to each $A_n$.  The prime elements of $A$ are exactly $V$, since any prime element in $A_n$ that was not a simple tensor gained a factorization in $A_{n+1}$, but any element of $V_n$ is prime in $A_n$ and remains prime in $A_m$ for all $m>n$.  Thus, every element of $A$ can be written uniquely as a simple tensor.
