The short answer is that the curl of the vector field you found is not zero, even at points where there is no current; so there cannot be a scalar potential for it. This is straightforward enough (if tedious) to verify: assign coordinates to the ends of the wire (I recommend $x = y= 0$ & $z = \pm d$); write out $\cos \alpha_1$ and $\cos \alpha_2$ in terms of $d$ and $\rho$, the distance from the axis (which is the same as your $a$); and take the curl of the resulting expression for $B_\phi$ in cylindrical coordinates. The $\rho$- and $z$-components of the result will be non-vanishing in general because $\partial (\rho B_\phi)/\partial \rho$ and $\partial B_\phi/\partial z$ are not zero.
As to why this happens, this is due to an underappreciated subtlety of the Biot-Savart Law.
For the Biot-Savart Law to yield a magnetic field satisfying Ampere's Law ($\nabla \times \mathbf{B} = \mu_0 \mathbf{J}$), it is necessary that $\nabla \cdot \mathbf{J} = 0$. Specifically, if you take the curl of $\mathbf{B}_\mathrm{BS}$ as defined by the Biot-Savart Law, then after heroic amounts of vector algebra (see, for example, §5.3.2 of Griffiths) you get to the statement that
$$
\left[ \mathbf{\nabla} \times \mathbf{B}_\mathrm{BS} \right]_\mathbf{r} = \mu_0 \mathbf{J}(\mathbf{r}) - \frac{\mu_0}{4 \pi} \iiint \left[\nabla_{\mathbf{r}'} \cdot \mathbf{J}(\mathbf{r'}) \right] \frac{\pmb{\mathscr{r}}}{\mathscr{r}^3} \, \mathrm{d}^3\mathbf{r'}
$$
where $\pmb{\mathscr{r}} \equiv \mathbf{r} - \mathbf{r}'$ and $\mathscr{r}$ is its magnitude.
For a current configuration which is divergence-free, this last term will vanish and everything's jake. However, in the case of a finite segment of wire, this is not the case; there is a "source" of current at one end of the wire, and a "sink" of current at the other. This means that you cannot expect this term to vanish, which means you cannot expect $\nabla \times \mathbf{B}_\mathrm{BS}$ to be zero. And if $\mathbf{B}_\mathrm{BS}$ is not curl-free in some region, there cannot be a scalar potential for $\mathbf{B}_\mathrm{BS}$ in that region.
To actually show that the curl of of $\mathbf{B}_\mathrm{BS}$ fails to vanish in this case, we can model the "source" and "sink" of current in this configuration via Dirac delta-functions:
$$
\nabla \cdot \mathbf{J} = 4 \pi I \left[ \delta^3(\mathbf{r} - \mathbf{r}_A) - \delta^3(\mathbf{r} - \mathbf{r}_B) \right],
$$
where $\mathbf{r}_A$ is the location of the "source" and $\mathbf{r}_B$ is the location of the "sink". If you plug these into the above integral, you find that
$$
\frac{\mu_0}{4 \pi} \iiint \left[\nabla_{\mathbf{r}'} \cdot \mathbf{J}(\mathbf{r'}) \right] \frac{\pmb{\mathscr{r}}}{\mathscr{r}^3} \, \mathrm{d}^3\mathbf{r'} = \mu_0 I \left( \frac{\mathbf{r} - \mathbf{r}_A}{|\mathbf{r} - \mathbf{r}_A|^3} - \frac{\mathbf{r} - \mathbf{r}_B}{|\mathbf{r} - \mathbf{r}_B|^3} \right)
$$
It is evident that the curl of $\mathbf{B}_\mathrm{BS}$ does not vanish anywhere in space; and of course this will be the same result you would get if you did the rigamarole described in the first paragraph. So there cannot exist a scalar potential such that $\mathbf{B}_\mathrm{BS} = - \nabla \Psi$.
You might also notice that the curl of $\mathbf{B}_\mathrm{BS}$ looks like an electric dipole field, and you might wonder why this is so. Well, in reality, if we have $\nabla \cdot \mathbf{J} \neq 0$, we would then have $\partial \rho /\partial t \neq 0$ as well. In this case, we would expect two point charges $\pm Q(t)$ at either end of the wire, creating a dipole electric field whose magnitude is increasing linearly with time. And the full time-dependent version of Ampere's Law is
$$
\nabla \times \mathbf{B} = \mu_0 \mathbf{J} + \epsilon_0 \mu_0 \frac{\partial \mathbf{E}}{\partial t}
$$
which means that at points not on the wire (where $\mathbf{J} = 0$), $\nabla \times \mathbf{B}$ will be proportional to a dipole electric field. Neat!
You may well wonder whether we could get around this by making the line segment into a closed loop (so that the "sources" and "sinks" cancel out), or making the finite segment into a infinite wire (so that the "source" and "sink" terms go to zero.) In this case, it is true that $\nabla \times \mathbf{B}_\mathrm{BS} = 0$ everywhere except at the points of the wire. However, the set of "all points in space minus the wire" will be topologically non-trivial in this case; specifically, it will not be simply connected. And the statement that "a curl-free vector field can always be written as the gradient of a scalar" is only true on simply connected spaces. So you will still be unable to find a scalar potential for $\mathbf{B}$.