I (only superficially) know that not every Lie algebra can be exponentiated to give a Lie group.
I also have only heard about the superconformal algebras, and not the superconformal groups. This is in contrast to the conformal algebras corresponding with the conformal group.
Does this illustrate the case of no Lie group being obtained by exponentiating the superconformal Lie algebra? And if this is the case, what happens when we try to exponentiate the superconformal algebra?
The physics literature uses the same terminology as you. But I have to start with this anyway just in case you aren't aware of it. The superconformal algebra should really be called "the superconformal superalgebra" because it is not a Lie algebra. Lie superalgebras are different in that their bilinear map is $\mathbb{Z}_2$ graded. That is, it acts as an anti-commutator on two odd elements and as a commutator otherwise.
Lie superalgebras are most often written as $\mathfrak{g}_0 \oplus \mathfrak{g}_1$ which are the even and odd parts respectively. The even part is a Lie algebra but the odd part on its own is neither a Lie algebra nor a Lie superalgebra. It is a representation of the even part which, if you want spin statistics, is required to be the spinor representation. If you have heard of the superconformal algebra containing a direct sum of Lie algebras, this is most likely a statement solely about $\mathfrak{g}_0$ which is a direct sum of the conformal algebra and the R-symmetry algebra (if there is one).
If your Lie superalgebra is finite dimensional (this only requires checking that its even part is finite dimensional) then it can be exponentiated to obtain a supergroup. A useful type of group element to consider is $$ g(x, \theta) = \exp \left ( x^\mu P_\mu + \theta^\alpha Q_\alpha \right ) \quad (1) $$ where $\theta$ is a Grassman co-ordinate. Checking how this acts on representations of the supergroup is not harder than the $\theta = 0$ case because superfields can be Taylor expanded as $$ \Phi(x, \theta) = \phi(x) + \theta^\alpha \psi_\alpha(x) + \dots \quad (2) $$ with a finite number of terms. In other words, you get to see how the component fields transform into each other under the supersymmetry action and therefore which combinations of superfields are invariant. It is also common to see people apply (1) to $\Phi(0, 0)$ and compare to (2) in order to see which component fields correspond to which superdescendants.
There is a dictionary which has more information on this even if the order of presentation is annoying. One idea it gets into is defining supergroups in terms of supermatrices where generalizations of the usual matrix operations apply. But, as with CFT, this structure would be hidden in SCFT due to the transformations that act nonlinearly in the physical space.
