Summarizing some comments, and continuing: the main point is that Haar measure is translation-invariant (and for non-abelian groups, in general, left-invariant and right-invariant are not identical, but the discrepancy is intelligible).
Unless you have intentions to do something exotic (say, on not-locally-compact, or not-Hausdorff "groups", which I can't recommend), you'll be happier later to have a regular measure, so, yes, the measure of a set is the inf of the measures of the opens containing it, and is the sup of the measures of the compacts contained in it, and, yes, the measure of a compact is finite. Probably you will also want completeness, especially when taking products, so subsets of measure-zero sets have measure zero.
Probably you'll want your groups to be countably-based, too, to avoid some measure-theoretic pathologies.
Then, for abelian topological groups (meaning locally compact, Hausdorff, probably countably-based), the basics of "Fourier series/transforms" work pretty well, as in Pontryagin and Weil. The non-abelian but compact case also turns out very well.
It seems the following.
A topological group $G$ is called $\omega$-precompact or $\omega$-bounded, if for each neighborhood $U$ of the unit there exist a countable subset $C$ of the group $G$ such that $G=CU=UC$ (or, equivalently, if for each neighborhood $U$ of the unit there exist a countable subset $C$ of the group $G$ such that $G=CU$).
It is obvious, that a locally compact topological group $G$ is $\sigma$-compact iff $G$ is $\omega$-precompact.
Proposition. Let $G$ be a locally compact group with a left invariant $\sigma$-finite and $\sigma$-additive measure $\mu$ on the Borel algebra $B(G)$ on the group $G$ such that $\mu(U)>0$ for each non-empty open subset $U$ of a space $G$ and there exists a non-empty open subset $U_0$ such that $\mu(U_0)<\infty$. Then the group $G$ is $\sigma$-compact.
Proof . Assume the converse. Then the group $G$ is not $\omega$-precompact. This easily implies that there exists a non-empty open subset $V$ of the group $G$ and an uncountable subset $A$ of the group $G$ such that the family $\{aV: a\in A\}$ is disjoint.$^*$ Let $v_0\in V$, $u_0\in U_0$ be an arbitrary points. Put $U=V\cap v_0u_0^{-1}U_0$. Then $0<\mu(U)<\infty$ and the family $\{aU: a\in A\}$ is disjoint. Let $\{G_n\}$ be a family of $\mu$-measurable subsets of the group $G$ such that $\mu(G_n)<\infty$ for each $n$ and $G=\bigcup G_n$. Then for each $n$ and each $a\in A$ the set $G_n\cap aU$ is $\mu$-measurable and has finite measure $\mu$. The additivity of the measure $\mu$ implies that for each $n$ and each $\varepsilon>0$ the set $\{a\in A:\mu(G_n\cap aU)>\varepsilon\}$ is finite.$^{**}$ Therefore the set $\{(n,a): \mu(G_n\cap aU)>0\}$ is countable. So if the set $A$ is uncountable then there exists an element $a\in A$ such that $\mu(G_n\cap aU) =0$ for each $n$. Since the measure $\mu$ is $\sigma$-additive, we have $\mu(aU)=\mu(U)=0$, a contradiction. $\square$
$^*$ This is a standard fact in the theory of topological groups. Since the group $G$ is not $\omega$-precompact, there
exists a neighborhood $U$ of the unit of $G$ such that $CU\ne G$ for each countable subset $C$ of the group $G$. Pick a neighborhood $V$ of the unit of $G$ such that $V=V^{-1}$ and $V^2\subset U$. From
here we can construct the required set $A$ by any of two ways.
On the first way we pick as $A$ a maximal subset of $G$ such that the family $\{aV: a\in A\}$ is disjoint. Such a subset $A$ exists by Zorn Lemma. We claim that $AU=G$. Indeed, assume the converse,
there exists a element $g\in G\setminus AU$. If $aV\cap gV\ne\varnothing$ for some $a\in A$, then
$g\subset aVV^{-1}\subset aU\subset AU$, a contradiction. Thus $aV\cap gV=\varnothing$ for each $a\in A$,
and a family $\{aV: a\in A\cup\{g\}\}$ is disjoint, which contradicts the maximality of $A$. Since $AU=G$, by the previuous paragraph, $A$ is uncountable.
On the first way, using the claim from the prepreviuous paragraph, by transfinite recursion we can construct $A$ as a sequence $A=\{a_\alpha:\alpha<\omega_1\}$ of elements of $G$ such that $a_\alpha\not\in a_\beta U$ for each $\alpha<\omega_1$ and each $\beta<\alpha$. This also implies that $a_\alpha V\cap a_\beta V=\varnothing$, which can be shown using the ideas from the previous paragraph.
$^{**}$ Indeed, assume that for some $n$ and $\varepsilon>0$ the set $\{a\in A:\mu(G_n\cap aU)>\varepsilon\}$ is infinite. Pick any sequence $\{a_m\}$ of distinct points of this set. Then $$\mu(G_n)\ge \mu(G_n\cap\{a_1,\dots, a_m\}U)=\sum_{i=1}^m \mu(G_n\cap a_iU)\ge m\varepsilon$$ for each $m$, which contradicts to $\mu(G_n)<\infty$.
Best Answer
This is to address only the case $p=\infty$.
Let $\lambda$ be a Haar measure (left invariant) on the locally compact Hausdorff topological group $G$. Let $f\in L_1(\lambda)$ and $g\in L_\infty(\lambda)$. Then $$f*g(x):=\int_G f(y)g(y^{-1}x)\lambda(dy)=\int_G f(xy)g(y^{-1})\,\lambda(dy)$$ If $\phi(y):=g(y^{-1})$, notice that $|\phi(y)|\leq\|g\|_{\infty}$ for $\lambda$-a.s $y\in G$ (the measure $\rho(A)=\lambda(A^{-1})$ preserves sets of measure zero, in fact $\rho(dy)=\triangle(y^{-1})\lambda(dy)$, where $\triangle$ is the modular function). From this, it follows that $$|f*g(x)|\leq\|g\|_\infty\int_G|f(xy)|\,\lambda(dy)=\|g\|_\infty\|f\|_1$$
Comment: With a little more effort one can probe that in fact $f*g\in\mathcal{C}(G)$.