Kinetics question regarding Rate Law
We can do this question in two steps, finding the rate law with respect to both A and $\ce{H+}$ individually.
$$\text{Rate}=[A]^y[\ce{H+}]^x$$

The first part is simple, they've given that the half life is independent of $[A]$, therefore it is first order with respect to $\ce{A}$.
Therefore as you've found,
$$\text{Rate}=[A][\ce{H+}]^x$$

Now, onto finding the order with respect to $\ce{H+}$.
From what I've seen on the contradictory answers on the net that OP has mentioned, they state that since the rate is inversely proportional to concentration of $\ce{H+}$, it is second order with respect to $\ce{H+}$ by the relation:
$$t_{1/2}\, \alpha\, [A]^{1-n}$$
where $n$ is the order with respect to A in a reaction.
Is this true? Yes. Is it correct in this context? No. This is because this is true only when the reactant in question actually decreases in the mixture, i.e.
$$-\frac{\mathrm d[A]}{\mathrm dt}=k[A]^n$$
Here, it is specifically given that it is in a buffer of $\mathrm{pH} = 4$. This means that there is no change in the actual concentration of $\ce{H+}$.
Therefore, what you've done is conceptually sound and is the correct method.
$$k'=k[\ce{H+}]^x = \text{constant}$$
\begin{align}
\frac{t_{1/2}(\mathrm{pH}=5)}{t_{1/2}(\mathrm{pH}=4)} &amp;= \frac{k'(\mathrm{pH}=4)}{k'(\mathrm{pH}=5)} = \frac{k[\ce{H+}]^x(\mathrm{pH}=4)}{k[\ce{H+}]^x(\mathrm{pH}=5)} \\
\implies \frac{100}{10}&amp;=\left(\frac{10^{-4}}{10^{-5}}\right)^x = 10^x \\
\implies x &amp;=1
\end{align}
Therefore, the rate would be:
$$\text{Rate}=k[A][\ce{H+}]$$