The classical example of a nonlinear Lie group is $\mathrm{SL}(2,\mathbb{R})$ (the group of 2-by-2 real matrices with determinant one).
But, are there any one-dimensional nonlinear Lie groups?
For example, consider $a,b\in\mathbb{C}$, equipped with an operation defined $$a*b\equiv\sqrt{a^2+b^2}.$$ It would seem that $\mathbb{C}$ forms a Lie group under $*$, with identity $e=0$. And it would seem to be nonlinear. But I suppose, because it acts on the complex plane, it's two-, not one-dimensional.
Here is how one can classify all connected 1-dimensional Lie groups $G$ over a field $F$ which is either ${\mathbb R}$ or ${\mathbb C}$.
The Lie algebra ${\mathfrak g}$ of $G$ is 1-dimensional, hence, commutative, hence, isomorphic to $F$ (with trivial Lie bracket structure). Let $\tilde{G}$ be the universal cover of $G$; it is a simply connected Lie group over $F$. It is a nice exercise to see that the exponential map is an isomorphism from ${\mathfrak g}\cong F$ (regarded as an abelian Lie group) to $G$. Hence, there is a unique simply-connected Lie group over $F$.
Now, $G$ itself is isomorphic to the quotient of $\tilde{G}$ by a discrete subgroup $L\subset \tilde{G}$. It is again a nice exercise to identify discrete subgroups of $(F, +)$:
In the real case, you only have $\{0\}$ and ${\mathbb Z}$. The quotient groups $G$ are isomorphic to (respectively) ${\mathbb R}$ and $S^1\cong U(1)$. Both are real-linear.
In the complex case, you have discrete subgroups isomorphic to $\{0\}, {\mathbb Z}$ and ${\mathbb Z}^2$.
Their quotient groups ${\mathbb C}/L$ are isomorphic, respectively, to ${\mathbb C}$, ${\mathbb C}^\times$ (both are complex-linear) and, as a real Lie group, to $S^1\times S^1$. the latter group is real-linear. However, it cannot be complex-linear, i.e. cannot be isomorphic to a complex Lie subgroup of $GL(n, {\mathbb C})$. The reason is the Liouville's theorem: Every holomorphic function on the compact connected complex manifold, such as a real torus $T^2$, is constant. Since the matrix coefficients are holomorphic functions, we obtain that the only compact connected complex Lie subgroups of $GL(n, {\mathbb C})$ is the trivial subgroup.