The ellipses is where I'm stuck. I don't think a simple algebraic manipulation will work here.
[Math] Showing when Young’s Inequality is in fact equality.
functional-analysismeasure-theoryreal-analysis
Related Solutions
This proof is from "Mathematical Toolchest" published by the Australian Mathematics Trust (image).
Example. If $p$ and $q$ are positive rationals such that $\frac1p + \frac1q = 1$, then for positive $x$ and $y$ $$\frac{x^p}p + \frac{y^q}q \ge xy.$$
Since $\frac1p + \frac1q = 1$, we can write $p = \frac{m+n}m$, $q = \frac{m+n}n$ where $m$ and $n$ are positive integers. Write $x = a^{1/p}$, $y = b^{1/q}$. Then $$\frac{x^p}p + \frac{y^q}q = \frac a{\frac{m+n}m} + \frac b{\frac{m+n}n} = \frac{ma + nb}{m + n}.$$
However, by the AM–GM inequality, $$\frac{ma + nb}{m + n} \ge (a^m \cdot b^n)^{\frac1{m+n}} = a^{\frac1p} b^{\frac1q} = xy,$$ and thus $$\frac{x^p}p + \frac{y^q}q \ge xy.$$
Let $1 \leqslant p \leqslant +\infty$. For $x\in \mathbb{R}$ [it works with the same argument in $L^p(\mathbb{R}^n)$ for any $n > 0$], let $\tau_x \colon L^p(\mathbb{R}) \to L^p(\mathbb{R})$ be given by
$$\tau_x(f) \colon u \mapsto f(u-x).$$
By the translation-invariance of the Lebesgue measure, $\tau_x$ is an isometry of $L^p(\mathbb{R})$ for every $x$.
Further, let $\rho \colon L^p(\mathbb{R}) \to L^p(\mathbb{R})$ be the reflection, $\rho(f) \colon u \mapsto f(-u)$. Then $\rho$ is also an isometry of $L^p(\mathbb{R})$ since the Lebesgue measure is also invariant under reflection.
Thus for any $f,g\in L^2(\mathbb{R})$ we can write
$$(f\ast g)(x) = \int_\mathbb{R} f(x-u)g(u)\,du = \int_\mathbb{R} \bigl(\rho(\tau_x(f))\bigr)(u)g(u)\,du.$$
For real-valued functions, we then see
$$(f\ast g)(x) = \langle \rho(\tau_x(f)), g\rangle_{L^2},\tag{1}$$
and for complex-valued functions, we need to conjugate one of the arguments of the inner product, and then note that complex conjugation is also an isometry of $L^p(\mathbb{R}^n)$ for all $1\leqslant p \leqslant +\infty$.
Now we can apply the Cauchy-Schwarz inequality to the right hand side of $(1)$ to obtain
$$\lvert (f\ast g)(x)\rvert \leqslant \lVert \rho(\tau_x(f))\rVert_2\cdot \lVert g\rVert_2\tag{2}.$$
But since $\rho$ and $\tau_x$ are isometries, the right hand side of $(2)$ is independent of $x$, and we have
$$\lvert (f\ast g)(x)\rvert \leqslant \lVert f\rVert_2\cdot\lVert g\rVert_2.\tag{3}$$
By $(3)$, the convolution $f\ast g$ is uniformly bounded, and taking the supremum over all $x\in\mathbb{R}$ yields
$$\lVert f\ast g\rVert_\infty = \sup_{x\in\mathbb{R}} \lvert (f\ast g)(x)\rvert \leqslant \lVert f\rVert_2\cdot\lVert g\rVert_2.\tag{4}$$
Since $f\ast g$ is continuous and $(f\ast g)(x) \to 0$ for $\lvert x\rvert \to \infty$ - note that these properties need not be assumed, they can be proved under the assumptions - the supremum in $(4)$ is actually attained, and can be replaced by $\max$ if one so wishes.
Basically the same argument, using the more general Hölder inequality instead of the Cauchy-Schwarz inequality shows that for $1\leqslant p,q\leqslant +\infty$ with $\frac{1}{p} + \frac{1}{q} = 1$ we have
$$\lVert f\ast g\rVert_\infty \leqslant \lVert f\rVert_p\cdot \lVert g\rVert_q$$
for all $f\in L^p(\mathbb{R})$ and $g\in L^q(\mathbb{R})$.
For $1\leqslant p < +\infty$, the continuous functions with compact support are dense in $L^p(\mathbb{R})$, and from that follows that for any fixed $f\in L^p(\mathbb{R})$ the function $x \mapsto \tau_x(f)$ is continuous, which shows that $f\ast g$ is continuous - if $p = \infty$, then $q = 1 < \infty$, and we note that we have
$$(f\ast g)(x) = \int_\mathbb{R} f(u)g(x-u)\,du,$$
so the continuity of $x\mapsto \tau_x(g)$ yields the continuity of $f\ast g$.
For $1 < p,q < +\infty$, we have
$$\lim_{\lvert x\rvert\to\infty} (f\ast g)(x) = 0,$$
as one can again see by using the denseness of $C_c(\mathbb{R})$ in $L^p(\mathbb{R})$ and $L^q(\mathbb{R})$ then, but that need not hold in the $\{ p,q\} = \{1,\infty\}$ case.
Best Answer
Here is an alternative argument for the equality condition:
Young's Inequality: $\forall p\in]1,\infty[, \frac{1}{p}+\frac{1}{q}=1,\forall a,b\in]0,\infty[: ab\leq\frac{a^p}{p}+\frac{b^q}{q}$. Furthermore
$$ab=\frac{a^p}{p}+\frac{b^q}{q} \iff a^p=b^q.$$
Observation: \begin{align} a^p=b^q \iff & a^{p/q}=b \iff & a^{p-1}=b \\ & a=b^{q/p} & a=b^{q-1} \end{align}
Lemma: Set
\begin{align} f:&]0,\infty[\to\Bbb{R} \\ &x\mapsto (1-x^{1-p})+(p-1)(1-x). \end{align} Then $f(x)=0 \iff x=1.$
Proof of Lemma: First observe that $f(1)=0$. Also $f'(x)=(p-1)(x^{-p}-1)$, from which we deduce that $1$ is the maximum point of $f$. $\checkmark$
Proof of the Equality Condition:
$(\impliedby)$ First suppose $a^p=b^q$. Then by the observation above
\begin{align} \dfrac{a^p}{p}+\dfrac{b^q}{q}=\dfrac{(q+p)b^q}{pq}=b^q=b^{q-1}b=ab. \end{align}
$(\implies)$ Conversely suppose we have equality. Then
\begin{align} &ab=\dfrac{a^p}{p}+\dfrac{b^q}{q} =\dfrac{qa^p+pb^q}{p+q}\\ &\implies p+q =q\dfrac{a^{p-1}}{b} + p\dfrac{b^{q-1}}{a}\\ &\implies q\left(\dfrac{a^{p-1}}{b}-1\right)=p\left(1-\dfrac{b^{q-1}}{a}\right)\\ &\implies \left(\dfrac{a^{p-1}}{b}-1\right)=\dfrac{p}{q}\left(1-\dfrac{b^{q-1}}{a}\right)\\ &\implies \left(\left(\dfrac{b^{q-1}}{a}\right)^{1-p}-1\right)=(p-1)\left(1-\dfrac{b^{q-1}}{a}\right)\\ &\implies 0=\left(1-\left(\dfrac{b^{q-1}}{a}\right)^{1-p}\right)+(p-1)\left(1-\dfrac{b^{q-1}}{a}\right) \end{align}
Hence $\dfrac{b^{q-1}}{a}$ is a root of $f$ in the above lemma, which guarantees that $\dfrac{b^{q-1}}{a}=1$. By the observation this is equivalent to $a^p=b^q$.
Note: I used certain arithmetical properties of the conjugates $(p,q)$. I'll leave those and the observation for you to check.