Is it true, that every prime (except 2) can be found as a divisor of enough long series of 1-s?

We can prove more very simply: if $\rm\:m\:$ is coprime to $10\,$ then any number with $\rm\: m\:$ digits all $\ne 0$ has a contiguous digit subsequence that forms a number divisible by $\rm\:m.\:$ Suppose the digits are $\rm\:d_{m}\ldots d_1.\:$ By $\rm\,d_i\ne 0\:$ the $\rm\:m\!+\!1\:$ numbers $\rm\:0,\,d_1,\, d_2 d_1,\, d_3 d_2 d_1,\, \ldots,d_m\!\ldots d_1$ are distinct. By Pigeonhole two are congruent $\rm\:mod\ m,\:$ so $\rm\:m\:$ divides their difference $\rm = 10^k\:$ times the number $\rm\,n\ne 0\,$ formed by the extra digits of the longest, so $\rm\:m\,|\,10^kn\:$ $\Rightarrow$ $\rm\:m\:|\:n,\:$ by $\rm\:m\:$ is coprime to $10.\:$

Let's do a simple example. Let $\rm\:m=9,\:$ and let the number be $\rm\:98765\color{green}{432}1.\:$ Modulo $9$ we have $\rm\:1 \equiv\color{#C00} 1,\ 21\equiv 3,\ 321\equiv 6,\ 4321\equiv\color{#C00} 1,\:$ so $\rm\:9\:|\:4321\!-\!1 = 432\cdot 10,\:$ so $\,9\,|\,\color{green}{432}.$

In your case, the divisor $\rm\:m=p\:$ is a prime coprime to $10$, so the number $\,111\ldots 111$ $\rm\,(m$ digits) does the trick, i.e. some subsequence $11\ldots 11$ is divisible by $\rm\:m.$

The result extends to any number having $\rm\:m\:$ nonzero digits: simply take the subsequences beginning with nonzero leading digit. This implies that the $\rm\:m+1\:$ numbers are increasing (so distinct), and the number formed by the extra digits is nonzero, since its leading digit is nonzero.