In general not. Consider the homomorphism $\mathrm{Aff}(p)\rightarrow\mathbb{F}_p^*$ mapping $ax+b$ to $a$. The probability that random elements generate a subgroup for which this map is surjective equals the probability that two random elements in $\mathbb{F}_p^*$ generate this group, which is $$\frac{1}{(p-1)^2}\sum_{d|p-1} \mu\left(\frac{p-1}{d}\right) d^2 = \sum_{t|p-1}\frac{\mu(t)}{t^2} = \prod_{q|p-1}\left(1-\frac{1}{q^2}\right),$$ where in the last product $q$ runs over primes only. Depending on the prime factors of $p-1$ this quantity is somewhere in the interval $[\frac{6}{\pi^2}, \frac{3}{4}]$.

On the other hand two random elements almost surely generate a subgroup containing a translation, and therefore all translations, thus the probability that two random elements generate $\mathrm{Aff}(p)$ is asymptotically equal to the quantity given above.

In general not. Consider the homomorphism $\mathrm{Aff}(p)\rightarrow\mathbb{F}_p^*$ mapping $ax+b$ to $a$. The probability that two random elements generate a subgroup for which the induced map is still surjective equals the probability that two random elements in $\mathbb{F}_p^*$ generate this group, which is $$\frac{1}{(p-1)^2}\sum_{d|p-1} \mu\left(\frac{p-1}{d}\right) d^2 = \sum_{t|p-1}\frac{\mu(t)}{t^2} = \prod_{q|p-1}\left(1-\frac{1}{q^2}\right),$$ where in the last product $q$ runs over primes only. Depending on the prime factors of $p-1$ this quantity is somewhere in the interval $[\frac{6}{\pi^2}, \frac{3}{4}]$.

On the other hand two random elements almost surely generate a subgroup containing a translation, and therefore all translations, thus the probability that two random elements generate $\mathrm{Aff}(p)$ is asymptotically equal to the quantity given above.

In general not. Consider the homomorphism $\mathrm{Aff}(p)\rightarrow\mathbb{F}_p^*$ mapping $ax+b$ to $a$. The probability that random elements generate a subgroup for which this map is surjective equals the probability that two random elements in $\mathbb{F}_p^*$ generate this group, which is $$\frac{1}{(p-1)^2}\sum_{d|p-1} \mu\left(\frac{p-1}{d}\right) d^2 = \sum_{t|p-1}\frac{\mu(t)}{t^2}.$$ Depending on the prime factors of $p-1$ this quantity is somewhere in the interval $[\frac{6}{\pi^2}, \frac{3}{4}]$.

On the other hand a random element almost surely generates a translation, thus the probability that two random elements generate $\mathrm{Aff}(p)$ is asymptotically equal to the quantity given above.

In general not. Consider the homomorphism $\mathrm{Aff}(p)\rightarrow\mathbb{F}_p^*$ mapping $ax+b$ to $a$. The probability that random elements generate a subgroup for which this map is surjective equals the probability that two random elements in $\mathbb{F}_p^*$ generate this group, which is $$\frac{1}{(p-1)^2}\sum_{d|p-1} \mu\left(\frac{p-1}{d}\right) d^2 = \sum_{t|p-1}\frac{\mu(t)}{t^2} = \prod_{q|p-1}\left(1-\frac{1}{q^2}\right),$$ where in the last product $q$ runs over primes only. Depending on the prime factors of $p-1$ this quantity is somewhere in the interval $[\frac{6}{\pi^2}, \frac{3}{4}]$.

On the other hand two random elements almost surely generate a subgroup containing a translation, and therefore all translations, thus the probability that two random elements generate $\mathrm{Aff}(p)$ is asymptotically equal to the quantity given above.