Previous answer was broken, here is what I can salvage.

To be clear, I am taking a projective representation to be a map $G \to PGL(V)$ with $\dim V \geq 2$. I call it irreducible if the corresponding representation $\tilde{G} \to GL(V)$, where $\tilde{G}$ is a central extension of $G$, is irreducible. I call two projective representations $G \to PGL(V)$ and $G \to PGL(W)$ isomorphic if I can choose an isomorphism $V \cong W$ to make them coincide. The goal is to classify $G$ such that $G$ only has one irreducible projective representation up to isomorphism.

Let $A=[G,G]$ and $B = G^{ab}$, so we have $1 \to A \to G \to B \to 1$. I can show

$A$ is elementary abelian (possibly trivial).

The conjugacy action of $B$ on $A \setminus \{ e \}$ is transitive.

$B$ is either of the form $C_m$ or $C_{2m} \times C_2$. (Here $C_k$ is the cyclic group of order $k$.)

This is a pretty limited list; I am not sure which groups on this list have the stated property.

So, let's prove this. Define $A=[G,G]$ and $B = G^{ab}$ and suppose $G$ has the claimed property.

**Lemma** All non-identity elements of $A$ are conjugate in $G$. (The old version of this answer said conjugate in $A$; that's false.)

**Proof** Suppose for the sake of contradiction that $A$ has $\geq 3$ conjugacy classes in $G$. Since the character table is invertible, there are two characters, $\chi_1$ and $\chi_2$ with $\chi_i$ coming from $\rho_i : G \to GL(V_i)$ which are linearly independent of each other and of the trivial character. Any one dimensional rep of $G$ is trivial on $A$, so $\dim V_1$ and $\dim V_2 \geq 2$.

By assumption, the projective representations $\bar{\rho}_i : G \to PGL(V_i)$ are isomorphic, choose an isomorphism $V_1 \cong V_2$ such that $\bar{\rho}_1 = \bar{\rho}_2$. Then $\rho_1(g) = \chi(g) \rho_2(g)$ for some $\chi: G \to \mathbb{C}^{\ast}$. It is easy to see that $\chi$ is a (one-dimensional) character of $G$. But then $\chi$ must be trivial on $A$, so $\chi_1 = \chi_2$ on $A$, contradicting our linear independence. $\square$

So the conjugacy action of $G$ on $A$ is transitive on $A \setminus \{ e \}$. By a standard lemma, this implies $A$ is elementary abelian. We've proved the first two bullet points.

Now, we study the structure of the abelian group $B$. Any irreducible projective representation of $B$ is an irreducible projective representation of $G$, so we deduce that $B$ has at most one irreducible
projective representation (none if $B$ is cyclic.)

Now, if $m>2$, then $C_m^2$ has multiple irreducible projective representations corresponding to different representations of the Heisenberg group of order $m^3$. (We can see this from the above argument: the commutator subgroup of the Heisenberg group is central, so the conjugacy action on the commutator subgroup is trivial and can't be transitive.) So $B$ must not have $C_m^2$ as a quotient for $m>2$. Also, $C_2^3$ has multiple irreducible representations, by projecting onto $C_2^2$ in different ways. Looking at the classification of finite abelian groups, this limits us to $C_m$ or $C_2 \times C_{2m}$.

I have not figured out which groups on the above list actually have the specified property.

I now have the list of candidates down to the following:

$C_m$ -- this has no irreducible projective representations, so you can decide whether to count it or not.

$C_{2m} \times C_2$. The only irreducible projective representation of this is through the quotient $C_2 \times C_2$, so it works.

$C_p^k \rtimes C_{(p^k-1)t}$, where the action is by a cyclic subgroup of $GL(\mathbb{F}_p)$ of order $p^k-1$. I have not yet figured out whether or not this works.

$G \times C_s$ where $s$ is odd and $G$ is a non-abelian central extension of the form $1 \to C_2 \to G \to C_{2^k} \times C_{2} \to 1$. I can show that all projective representations of this factor through the quotient $G$, but I haven't figured out for which $G$ there is a unique irreducible projective representation.

I'll try to add proofs of what I have later today.