In principle it is possible to determine every (edit: connected) Lie group having Lie algebra some finite-dimensional real Lie algebra $\mathfrak{g}$. As radekzak says in the comments, first we find the unique simply connected Lie group $\widetilde{G}$ with Lie algebra $\mathfrak{g}$, and then every other Lie group $G$ with Lie algebra $\mathfrak{g}$ must admit $\widetilde{G}$ as its universal cover. Standard arguments can be used to show that the covering map $\widetilde{G} \to G$ is given by quotienting by a discrete central subgroup of $\widetilde{G}$, which can be identified with $\pi_1(G)$; conversely any such quotient gives a Lie group with Lie algebra $\mathfrak{g}$.
The conclusion is that (edit: connected) Lie groups with Lie algebra $\mathfrak{g}$ are classified by discrete central subgroups of $\widetilde{G}$ (up to the action of the automorphism group $\text{Aut}(\widetilde{G}) \cong \text{Aut}(\mathfrak{g})$). For classical Lie algebras it's usually not hard to identify $\widetilde{G}$ and its center explicitly; for example if $\mathfrak{g} = \mathfrak{su}(n)$ then $\widetilde{G} = SU(n)$, whose center is the subgroup $\mu_n$ of $n^{th}$ roots of unity. The conclusion is that the Lie groups with Lie algebra $\mathfrak{su}(n)$ are precisely the quotients $SU(n)/\mu_k$ where $k \mid n$. Most of these are not very familiar Lie groups except for $SU(2)/\mu_2 \cong SO(3)$ and maybe $SU(n)/\mu_n \cong PSU(n)$.
You might not consider this completely satisfying since it's not clear a priori how to find $\widetilde{G}$ given $\mathfrak{g}$ in general. But this isn't so bad: using Levi decomposition we can reduce this problem to the case that $\mathfrak{g}$ is semisimple and the case that $\mathfrak{g}$ is solvable. The case that $\mathfrak{g}$ is semisimple can be done explicitly using the classification of semisimple Lie algebras and I am pretty sure we can explicitly compute the center in all of these cases.
The case that $\mathfrak{g}$ is solvable is more annoying. In principle if we write $\mathfrak{g}$ as an iterated extension of abelian Lie algebras then $\widetilde{G}$ is correspondingly determined as an iterated extension of abelian Lie groups but this seems messy. Alternatively, after a single extension $0 \to [\mathfrak{g}, \mathfrak{g}] \to \mathfrak{g} \to \mathfrak{g}/[\mathfrak{g}, \mathfrak{g}] \to 0$ we can pass to the derived subalgebra $[\mathfrak{g}, \mathfrak{g}]$ which is nilpotent. For nilpotent Lie algebras the Baker-Campbell-Hausdorff formula terminates so can be used to explicitly write down a polynomial multiplication on $\mathfrak{g}$ making it isomorphic to $\widetilde{G}$.
Computing the center in general could get tricky. The issue is that we know that $Z(\widetilde{G})$ has Lie algebra $Z(\mathfrak{g})$, which we can compute in principle, but $Z(\widetilde{G})$ may not be connected. I don't know if there's a nice systematic way to figure out the elements of $Z(\widetilde{G})$ not connected to the identity in general.
Re: your comment about integrating the Lie algebra, this can actually be done: there is a uniform definition of $\widetilde{G}$ in terms of a certain space of paths in $\mathfrak{g}$ ("Lie integration"). I don't think it's very easy to use for computations, though, since it constructs $\widetilde{G}$ as a quotient of a messy infinite-dimensional thing; as far as I know it has only theoretical applications.
So, in principle it's possible to do this up to perhaps some difficulty computing the center. In typical cases $\mathfrak{g}$ is probably semisimple or reductive and in this case one can appeal to the classification of such Lie algebras and their corresponding Lie groups to compute the center explicitly.
