2 Further explanation with added references.

No. The simplest example I can think of is that $SO(4)$ is homeomorphic to $SO(3)\times Sp(1)$ as topological spaces, but they are not isomorphic Lie groups. In fact, there is a double covering $Sp(1)\times Sp(1)\to SO(4)$, $(q_1,q_2)\cdot x = q_1xq_2^{-1}$, where $Sp(1)$ is viewed as the unit quaternions and $x\in\mathbf H\cong\mathbf R^4$, which gives a Lie group isomorphism $Sp(1)\times Sp(1)/{\pm1}\cong SO(4)$.

On the other hand, every continuous homormorphism homomorphism between Lie groups is automatically smooth, so if the homeomorphism is a homomorphism, then it must be a Lie group isomorphism.

Edit: I was puzzled by the fact that all known examples are given by pairs of locally isomorphic groups. Then I found the following papers:

Toda, H., A note on compact semi-simple Lie groups, Japan J. Math. 2 (1976), 355-358.

in which he proves that "two simply connected, compact (and hence semi-simple) Lie groups are isomorphic to each other if and only if they have isomorphic homotopy groups for each dimension". Later this was generalized by S. Boekholt (Journal of Lie Theory Volume 8 (1998) 183-185) as follows:

"Let $G_1$ and $G_2$ be two compact, connected Lie groups with isomorphic homotopy groups in each dimension. Then $G_1$ and $G_2$ are locally isomorphic."

So, going back to the question, we cannot avoid local isomorphism.

1

No. The simplest example I can think of is that $SO(4)$ is homeomorphic to $SO(3)\times Sp(1)$ as topological spaces, but they are not isomorphic Lie groups. In fact, there is a double covering $Sp(1)\times Sp(1)\to SO(4)$, $(q_1,q_2)\cdot x = q_1xq_2^{-1}$, where $Sp(1)$ is viewed as the unit quaternions and $x\in\mathbf H\cong\mathbf R^4$, which gives a Lie group isomorphism $Sp(1)\times Sp(1)/{\pm1}\cong SO(4)$.

On the other hand, every continuous homormorphism between Lie groups is automatically smooth, so if the homeomorphism is a homomorphism, then it must be a Lie group isomorphism.