The answer is yes (in a certain sense; see below). This proof seems a bit long... so perhaps someone can come up with an easier argument!

The assumption that the sets form cliques in a graph seems to be irrelevant here. In fact, this problem isn't really a "graph problem" at all, so let's rephrase it in terms of hypergraphs. We want to show that there is unique (up to isomorphism) $(n-1)$-uniform hypergraph $\mathcal{H}$ such that

$\mathcal{H}$ has exactly $n$ edges,

any two edges of $\mathcal{H}$ intersect on exactly one vertex, and

for every vertex $v$, there exists a pair $e,e'$ of hyperedges with $e\cap e'=\{v\}$.

**Claim 1:** $\mathcal{H}$ has at most $\binom{n}{2}$ vertices.

*Proof:* Consider the function which maps each vertex $v$ to a pair of hyperedges whose intersection is $\{v\}$. This map is injective. $\square$

Okay, now order the edges of $\mathcal{H}$ by $e_1,\dots,e_n$ in an arbitrary order and let $\mathcal{H}_k$ be the subhypergraph of $\mathcal{H}$ induced by the hyperedges $e_1,\dots,e_k$. Clearly, $\mathcal{H}_1$ consists of $n-1$ vertices contained in a single hyperedge.

**Claim 2:** For $k\geq2$ we have $|V(\mathcal{H}_k)|\geq |V(\mathcal{H}_{k-1})| + n-k$.

*Proof:* By assumption, $e_k$ intersects each of the edges $e_1,\dots,e_{k-1}$ on exactly one vertex. Therefore, there must be at least $(n-1)-(k-1) = n-k$ vertices of $e_k$ not contained in any of the edges $e_1,\dots,e_{k-1}$. This completes the proof. $\square$

So, by Claim 2,

$$|V(\mathcal{H})| \geq |V(\mathcal{H}_1)| + (|V(\mathcal{H}_2)|-|V(\mathcal{H}_1)|) + \dots +(|V(\mathcal{H}_n)|-|V(\mathcal{H}_{n-1})|)$$
$$= (n-1) + (n-2)+\dots+1+0 = \binom{n}{2}.$$
Combining this with Claim 1, we see that we must have $|V(\mathcal{H})|=\binom{n}{2}$. Moreover, it must be the case that
$$|V(\mathcal{H}_k)| = \binom{n}{2} - \binom{n-k}{2}$$
for $0\leq k\leq n$ (otherwise, we could apply Claim 2 to obtain $|V(\mathcal{H})|>\binom{n}{2}$, contradicting Claim 1).

In particular, we get that any vertex $v$ is contained in exactly two hyperedges of $\mathcal{H}$ (otherwise, if we let $e_1,e_2,e_3$ be the three hyperedges containing $v$, then this contradicts the fact that $|V(\mathcal{H}_3)|=3n-6$).

Now, let us show that $\mathcal{H}_k$ is uniquely determined (up to isomorphism) for each $k$ by induction. The result is trivial for $k=1$. Now for $k\geq2$, we can assume (by induction) that $\mathcal{H}_{k-1}$ contains no vertex of degree greater than two and that each hyperedge contains exactly $n-k+1$ vertices of degree one. By the above arguments, the graph $\mathcal{H}_k$ must be constructed from $\mathcal{H}_{k-1}$ by

(a) adding a set $A$ of $n-k$ new vertices, and

(b) adding a new hyperedge $e_k$ containing $A$ and containing exactly one vertex $v_i$ of degree one from the hyperedge $e_i$ for $1\leq i\leq k-1$.

All that remains is to show that any such choice yields the same hypergraph, up to isomorphism. For $1\leq i\leq k-1$, let $u_i$ and $v_i$ be (possibly different) vertices of $e_i$ of degree one. There is an isomorphism from the hypergraph obtained by adding the edge $A\cup \{u_1,\dots,u_{k-1}\}$ and the hypergraph obtained by adding the edge $A\cup \{v_1,\dots,v_{k-1}\}$ where $v_i\mapsto u_i$, $u_i\mapsto v_i$ and every other vertex simply maps to itself. This completes the proof.

**Edit:** As others have pointed out, what I have proved is that these graphs are unique under the additional assumption that *every edge of $G$ is contained in one of the cliques $S_1,...,S_n$.* Without this assumption, the class of graphs is closed under adding edges, and therefore we cannot have uniqueness. I originally interpreted the problem as having this assumption, but I do agree that it was not included in the statement.