Topological semi-direct products In Kaniuth, Taylor, Induced representations of locally compact groups on pages 9-10 it's claimed that if $G$ is a locally compact group with closed subgroups $N,H$, with $N$ normal in $G$, with $N\cap H=\{e\}$, and with $NH=G$, then $G$ is a topological semidirect product of $N$ and $H$.
We copy the algebraic proof, defining an action $\alpha_h(n) = hnh^{-1}$ which will be suitably continuous, allowing us to construct $N \rtimes_\alpha H$.  The map $N \rtimes_\alpha H \rightarrow G; (n,h) \mapsto nh$ is an isomorphism of groups, and clearly continuous.

Why is the inverse of this map continuous?

You would need to show that given nets $(n_i)\subseteq N, (h_i)\subseteq H$ with $n_ih_i\rightarrow e$, then necessarily $n_i\rightarrow e, h_i\rightarrow e$.  I don't see how to do this.
(Under some conditions, e.g. that $N \rtimes_\alpha H$ is $\sigma$-compact, there are open mapping theorems for locally compact groups available, which would show this.  For example, see Corollary 1.7 in Hofmann, Morris, Open Mapping Theorem for Topological Groups (pdf).)
 A: Added: Below is my previous attempt at an answer.
It seems for $G$ be to the internal topological semidirect product of $N$ and $H$ with $H$ acting on $N$ via conjugation by definition $G$ ought to have the product topology $N\times H$. The continuity of $\pi: G\to H$ I (mistakenly) use below is in fact equivalent to this (see e.g. Bourbaki's General Topology, Chapters 1-4, p.242.).
In general, if $G$ is a topological group, $N,H\leq G$ are subgroups with N normal such that $NH=G$ and $N\cap H=1$ (so that $G\cong N\rtimes^{\text{Alg}} H$ as groups without specified topologies), then one can form the external topological semidirect product $N\rtimes^{\text{Top}} H$. Then the map $N\rtimes^{\text{Top}} H\to G, (n,h)\mapsto nh$ is always an isomorphism of groups and is continuous, but its inverse need not be continuous (see again Bourbaki).
For an example of an algebraic internal semidirect product that is not a topological internal semidirect product see Roelke & Dierolf's Uniform Structures on Topological Groups and Their Quotients, p.121, Ex.6.18.a.
It remains to be seen if one assumes additionally that $G$ is locally compact and $N,H$ are closed one obtains that $G$ has the product topology automatically (which is what was asked originally). As of the writing this latest edit I have not been able to produce a counterexample or give a proof. (In the Roelke-Dierolf counterexample $N$ is dense and index $2$ it seems.)
I will add further details if I make any progress.

For any $g\in G$, there are unique $n_g\in N$ and $h_g\in H$ such that $g=n_g h_g$. $\pi:G\to H, g\mapsto h_g$ is a surjective topological group homomorphism with kernel $\ker(\pi)=N$. If $n_\bullet h_\bullet \to e$, then $h_\bullet=\pi(n_\bullet h_\bullet)\to \pi(e)=e$, hence also $h_\bullet^{-1}\to e$. Thus $n_\bullet=n_\bullet h_\bullet h_\bullet^{-1}\to e$.
