If a right adjoint to the product functor exists, must it be the diagonal?

For the first question: Yes.

Let $C$ be a category with finite products and let $\Delta=(\Delta_1,\Delta_2):C\to C\times C$ s.t. $$\times\dashv \Delta.$$ More specifically we find $$[X\times Y, Z]\simeq[(X,Y),\Delta Z]=[X,\Delta_1 Z]\times[Y,\Delta_2 Z].$$ Note that this isomorphism is functorial in all arguments and given $f:X'\to X$ and $g:Y'\to Y$ we find the isomorphism to be compatible with precomposition with $(f,g)$ (I did not check this compatibility with much diligence so please take care here). Now, take $Y=*$ to be the final object (also the neutral object with respect to $\times$). We find $$[X,Z]\simeq[X,\Delta_1 Z]\times [*,\Delta_2 Z].$$ Taking $X=Z$ and we find $\varphi\in [Z,\Delta_1 Z]$ and $\omega\in [*,\Delta_2 Z]$ corresponding to the identity on $Z$. Conversely, taking $X=\Delta_1 Z$ we find $\phi\in[\Delta_1, Z]$ corresponding to $(\mathrm{id}_{\Delta_1 Z},\omega)$. Now the compatibility with precomposition can be used to obtain $$\varphi\circ\phi=\mathrm{id}\textrm{ and } \phi\circ\varphi=\mathrm{id}.$$

I guess you'd still have to confirm that this actually constitutes an isotransformation $\mathrm{id}_C\simeq\Delta_1$.

Edit: After this, I guess we also find $*$ to not only be final but also initial as we arrive at $$(-,\omega):[X,Z]\simeq[X,Z]\times[*,Z].$$ So $[*,Z]=\{\omega\}$.

Building on Garlef Wegart's work, the answer is yes in general. The idea is to show that if $C$ has a right adjoint to $\times$, then it is enriched in pointed sets, so that $C$ embeds fully faithfully in a category $C_\ast$ with a zero object and the easy argument can be applied. And as noted already, in this case $C$ will have biproducts with respect to a (necessarily unique) enrichment in pointed sets.

Suppose that $R= (R_1, R_2): C \to C \times C$ is right adjoint to $\times : C \times C \to C$. Then $\pi_1 : X \times Y \to X$ is natural in $X$ and $Y$. By adjointness, this corresponds to $\pi_{11}: X \to R_1 X$ and $\pi_{12}: Y \to R_2 X$ natural in $X$ and $Y$. Similarly, $\pi_2: Y \times X \to X$ gives us $\pi_{21} : Y \to R_1 X$ and $\pi_{22}: X \to R_2 X$. Thus we get $(\pi_{12}, \pi_{21}): Y \to R_2 X \times R_1 X$ natural in $X$ and $Y$, which we can compose (after swapping the factors) with the counit $R_1 X \times R_2 X \to X$ to obtain a map $Y \to X$, natural in $X$ and $Y$.

That is, $C$ admits a (necessarily unique) enrichment in pointed sets. So there is a canonical way to add a zero object, and the resulting inclusion $C \to C_\ast$ is fully faithful. Moreover, $C_\ast$ still has products, computed as in $C$ with $X \times 0 = 0 \times X = X$. One can extend $R$ by hand to a functor $R_\ast: C_\ast \to C_\ast \times C_\ast$ by setting $R_\ast(0) = (0,0)$, and then check by hand that $R_\ast$ is still right adjoint to $\times_{C_\ast}$. (Conceptually, zero objects are absolute for $\mathsf{Set}_\ast$-enrichment, and the forgetful functor from $\mathsf{Set}_\ast$-enriched categories to $\mathsf{Set}$-enriched categories creates products and adjunctions.)

Hence the argument I gave in the comments applies: $X \times Y = (X + 0) \times (0 + Y) = (X \times 0) + (0 \times Y) = X + Y$ in $C_\ast$, because $\times$, as a left adjoint, must preserve colimits (note that the coproduct $(X,0) + (0,Y) = (X,Y)$ holds automatically in $C_\ast$; we need not assume the existence of coproducts). Hence $\times_{C_\ast} = +_{C_\ast}$ and taking adjoints we have $R_\ast = \Delta_{C_\ast}$. By restricting along the fully faithful inclusion $C \to C_\ast$, we get $R= \Delta_C$ as desired.