I'm not quite sure if this is the answer that you looking for but anyway he we go. For a locally compact group you are going to generally want to look at strongly continuous representation. By this is mean endow $B(H)$, the bounded operators on a hilbert space $H$ with the topology of point-wise norm convergence. And only consider reps $\pi:G\rightarrow B(H)$ that are continuous with this topology. Now such a rep is unitary if, for every $g\in G$, $\pi(g)$ is a unitary operator.
Now the notion of "occur in" that you mention seems to be the notion of strong containment. We say that $\rho:G\rightarrow B(K)$ is strongly contained in $\pi:G\rightarrow B(H)$ if there is a $G$-equivarient unitary operator from $K$ to a closed subspace of $H$.
So it now seems that you are asking when does the left regular rep ($L^2(G)$) strongly contain all irreducibles. So yes for compact Lie groups this follows from Peter-weyl this is true.
However as soon as you go to something non-compact this might not be true.
In fact, there is a much weaker notion known as weak containment of representation. and it is known that $L^2(G)$ weakly contains all irreducible reps if and only if $G$ is amenable.
Non-compact Lie groups are in general not amenable, (any groups which contains $\mathbb{F}_2$ the free group on 2 generators is non-amenable)
There is much more to be said about this but I think that this should suffice for now
You can mimic the standard finite group argument. Recall that, if $X$ and $Y$ are two finite-dimensional representations of $G$, with characters $\chi_X$ and $\chi_Y$, then $\dim \mathrm{Hom}_G(X,Y) = \int_G \overline{\chi_X} \otimes \chi_Y$, where the integral is with respect to Haar measure normalized so that $\int_G 1 =1$. The proof of this is exactly as in the finite case.
Now, let $W=1 \oplus V \oplus \overline{V}$. Your goal is to show that, for any nonzero representation $Y$, $\mathrm{Hom}(W^{\otimes N}, Y)$ is nonzero for $N$ sufficiently large. So you need to analyze $\int (1+\chi_V + \overline{\chi_V})^N \chi_W$ for $N$ large. Just as in the finite group case, we are going to split this into an integral near the identity, and an exponentially decaying term everywhere else.
I'm going to leave a lot of the analytic details to you, but here is the idea. Let $d=\dim V$. The function $f:=1+\chi_V + \overline{\chi_V}$ makes sense on the entire unitary group of $V$, of which $G$ is a subgroup. On a unitary matrix with eigenvalues $(e^{i \theta_1}, \cdots, e^{i \theta_d})$, we have $f=1+2 \sum \cos \theta_i$. In particular, for an element $g$ in the Lie algebra of $G$ near the identity, we have
$$f(e^g) = (2n+1) e^{-K(g) + O(|g|^4)}$$
where $K$ is $\mathrm{Tr}(g^* g) = \sum \theta_i^2$.
And, for $g$ near the identity, $\chi_X(e^g) = d + O(|g|)$. So the contribution to our integral near the identity can be approximated by
$$\int_{\mathfrak{g}} \left( (2n+1) e^{-K(g) + O(|g|^4)} \right)^N (d+ O(|g|) \approx (2n+1)^N d \int_{\mathfrak{g}} e^{-K(\sqrt{N} g)} = \frac{(2n+1)^N d}{N^{\dim G/2}} C$$
where $C$ is a certain Gaussian integral, and includes some sort of a factor concerning the determinant of the quadratic form $K$. You also need to work out what the Haar measure of $G$ turns into as a volume form on $\mathfrak{g}$ near $0$, I'll leave that to you as well.
For right now, the important point is that the growth rate is like $(2n+1)^N d$ divided by a polynomial factor. In the finite group case, that polynomial is a constant, which makes life easier, but we can live with a polynomial.
Now, look at the contribution from the rest of $G$. For any point in the unitary group $U(d)$, other than the identity, we have $|f| < 2n+1$. (To get equality, all the eigenvalues must be $1$ and, in the unitary group, that forces us to be at the identity.) So, if we break our integral into the integral over a small ball around the identity, plus an integral around everything else, the contribution of every thing else will be $O(a^N)$ with $a<2n+1$.
So, just as in the finite group case, the exponential with the larger base wins, and the polynomial in the denominator is too small to effect the argument. Joel and I discussed a similar, but harder, argument over at the Secret Blogging Seminar.
Best Answer
The connected Lie groups whose points are separated by the finite-dimensional complex representations are exactly the linear Lie groups, for instance by Th. 5.3 in Beltiţă and Neeb - Finite-dimensional Lie subalgebras of algebras with continuous inversion.