A sketch of an argument (I'll leave it to you to figure out the precise conditions and formulations that makes this argument work):

Since $\Phi$ is even, let us just focus on the problem restricted to $\mathbb{R}_+$.

Write $\Psi(s) = \Phi\circ\exp(s)$ and $g_t(s) = f_t\circ \exp(s)$ for $s\in \mathbb{R}$. The problem now reduces asking about $g_t$ such that
$$ \lim_{t\to\infty} \int_{-\infty}^{\infty} \Psi(u-s) g_t(s) e^s ~\mathrm{d}s = \Psi(u) $$

We know a few things:

- $g_t$ is positive, hence also is $g_t(s) e^s$
- $\Psi$ is positive.
- $\Psi$ is strictly decreasing
- $\lim_{s\to -\infty} \Psi(s) = 1$
- $\Psi$ is concave on $\mathbb{R}_-$, and convex on $\mathbb{R}_+$.

**Assumption** $e^s g_t(s)\in L^1$. (Equivalently $f_t \in L^1$.), and $s e^s g_t(s) \in L^1$.

Then condition 4 and the equation implies that for sufficiently large $t$, $g_t$ has mass approximately 1.

On the other hand, condition 5 tells you that for sufficiently large $+ u$ translations you have that $$\int_{-\infty}^\infty \Psi(u-s) g_t(s) e^s \mathrm{d}s \geq \Psi(u - m_t) $$
where $m_t$ is the center of mass of $g_t(s) e^s$. The sign is reversed if we look at sufficiently large $+u$ translations. Using that $\Psi$ is strictly decreasing one of the two asymptotic behaviour contradicts $m_t \neq 0$.

With $m_t = 0$ the strict convexity/concavity near the infinities imply that the equation can only be satisfied if the support of $g_t(s) e^s$ is very concentrated near the set $\{0\}$, for $t$ sufficiently large. And a few more lines should give you that $g_t(s) e^s$ converges to the Dirac measure at 0.

Note, however, to get the statement that $f_t$ converges to the dirac support at 1, that one of the conditions that you need to impose would be decay rate on $g_t(s)$ as $s \to -\infty$, which translates also to some statement about behavior of $f_t$ near the origin. This is due to the following fact which poses another obstruction to the result you want in your question.

Let $h_t = 2t \mathbf{1}_{[1/(2t), 1/t]}$ so that $\int h_t = 1$.

Considering now that
$$\lim_{t \to \infty} \int_{-\infty}^\infty \Phi(u/\xi) h_t(\xi) \mathrm{d}\xi = 0 $$
but that $h_t$ does not converge in $L^1$, and that $h_t$ converges in the sense of distributions to $\delta_0$.