I see you've found an answer, but I'll just post what I had been writing up anyway.
Let $G$ be a group of order 1805. By applying Sylow's Theorems, we can tell that $G$ must have a normal Sylow 19-subgroup, call it $N$. Then we can write $G = NH$, where $H$ is any subgroup of order 5.
Now $H$ acts on $G$ by conjugation, and since $N$ is normal, this action restricts to an action on $N$. We can check to see what can happen depending on the isomorphism type of $N$. As a group of order $19^2$, $N$ is either $\mathbb Z_{19^2}$ or $\mathbb Z_{19} \times \mathbb Z_{19}$.
If $H$ acts trivially on $N$, then we have either $G \cong \mathbb Z_{5} \times \mathbb Z_{19^2}$ or $G \cong \mathbb Z_{5} \times\mathbb Z_{19} \times \mathbb Z_{19}$, which are the cases you've already found.
Now let's see which kinds of nontrivial actions could be happening.
If $N \cong \mathbb Z_{19^2}$, then $\operatorname{Aut}(N) \cong (\mathbb Z_{19^2})^\times$, which has order $\varphi(19^2) = 342$, which is not divisible by 5, so we can't have a nontrivial action in this case.
If $N \cong \mathbb Z_{19} \times\mathbb Z_{19}$, then $N$ is a 2 dimensional vector space over the field $\mathbb F_{19}$, and so $\operatorname{Aut}(N) \cong \operatorname{Gl}_2(\mathbb F_{19})$, which has order $(19^2-1)(19^2-19) = 2^4 \cdot 3^4 \cdot 5 \cdot 19$. Thus if $H$ has a nontrivial action on $N$, then it acts like a Sylow 5-subgroup of $\operatorname{Aut}(N)$. Since these subgroups are all conjugate and are cyclic, any two nontrivial actions of $H$ on $N$ will give isomorphic groups. This fact can be found on page 184 of Dummit and Foote's Abstract Algebra.