Are there some techniques for checking whether a statement implies another without truth tables?

Yes. I always think of using truth tables as a last resort; sometimes it actually pays off because the effort spent constructing the truth table can be considerably less than that using logical deductions, but there is not much at all learned by proving something with a truth table--it's very mechanical.

When trying to prove an implication of the form $\Omega\to\Psi$ without using a truth table, your job is to show that, given $\Omega$, the implication $\Omega\to\Psi$ is a tautology. I will give you two examples of how to check whether or not a statement implies another without a truth table (the first will not be a tautology but the second will be).

Example 1: Given $p$, does $p\to(p\land q)$?

Solution. Notice the following: \begin{align} p\to(p\land q) &\equiv \neg p\lor(p\land q)\tag{since $r\to s\equiv \neg r\lor s$}\\[0.5em] &\equiv \underbrace{(\neg p\lor p)}_{\text{Always true}}\;\;\land\underbrace{(\neg p\lor q)}_{\substack{\text{True if $q$ is true}\\\text{False if $q$ is false}}}\tag{distributivity} \end{align} Hence, the implication $p\to(p\land q)$ is not a tautology since it will be false when $q$ is false.

Example 2: Given $p\land q$, does $(p\land q)\to p$?

Solution. Notice the following: \begin{align} (p\land q)\to p &\equiv \neg(p\land q)\lor p\tag{since $r\to s\equiv \neg r\lor s$}\\[0.5em] &\equiv (\neg p\lor\neg q)\lor p\tag{DeMorgan}\\[0.5em] &\equiv \underbrace{(\neg p\lor p)}_{\text{Always true}}\lor\neg q.\tag{associativity of $\lor$} \end{align} Since $\neg p\lor p$ is always true, we have that $(p\land q)\to p$ is a tautology, and we proved this without using a truth table.