Let $(P_t)_{t\ge0}$ a convolution semigroup, i.e. a family of probability measures with the property $\forall s,t\ge 0: P_t\ast P_s=P_{t+s}$.
I want to show that $P_t\rightarrow \delta_0$ for $t\rightarrow 0$ with respect to weak convergence. Suppose $P_t\rightarrow Q\ne\delta_0$ for $t\rightarrow 0$. It follows
$$ P_{t}=P_{t/2}\ast P_{t/2}\rightarrow Q\ast Q$$
for $t\rightarrow 0$. Therefore $Q\ast Q =Q$. Now this is only true for $\delta_0$, which contradicts the assumption. Is this proof right?
Assuming you are talking about probability measures on the reals, and convolution with respect to $(\mathbb R, +)$, the answer is No. A convolution semigroup need not be continuous anywhere.
Let $f$ be any discontinuous solution of Cauchy's functional equation $f(x+y)=f(x)+f(y)$ (these are discontinuous everywhere, and unbounded in every neighborhood) and let $P_t=\delta_{f(t)}$ be the point mass concentrated at $f(t)$. Then the convolution property $P_t*P_s=P_{s+t}$ is satisfied, but the map $t\mapsto P_t$ is not weak* continuous. For if it were, since $x\mapsto \tanh(x)$ is continuous and bounded, the function $t\mapsto \int_{\mathbb R} \tanh( x)P_t(dx) = \tanh(f(t)) $ would have to be continuous. Which it is not, not even at $t=0$.
A flaw in your argument is assuming that the only way $P_t\to\delta_0$ can fail is for $P_t\to Q$ for some $Q\ne \delta_0$.