Let $(P,\preccurlyeq_P)$ and $(Q,\preccurlyeq_Q)$ be two posets. They are isomorphic if there exists a bijective map $f:P\to Q$ such that for all $a,b\in P$, $a\preccurlyeq_P b\Leftrightarrow f(a)\preccurlyeq_Q f(b)$.
If the poset $P$ is generated by the covering relation $\lessdot_P$ and $Q$ is generated by the covering relation $\lessdot_Q$, is it true that $P$ and $Q$ are isomorphic if there exists a bijective map $f:P\to Q$ such that for all $a,b\in P$, $a\lessdot_P b\Leftrightarrow f(a)\lessdot_Q f(b)$.
I'm not sure what you mean with $P$ being generated by the covering relation $\lessdot$.
To me, it makes sense if it means that $$a\leq b \Longleftrightarrow \exists c_1,\ldots,c_n \bigl( a=c_1\lessdot c_2 \lessdot \cdots \lessdot c_n = b \bigr).$$ In that case, in answer is yes.
Suppose that $f:P\to Q$ is a bijection satisfying $a\leq b$ iff $f(a)\leq f(b)$.
Now, if $a\lessdot b$, then $a\leq b$, whence \begin{align} a\lessdot b &\Longrightarrow f(a)\leq f(b)\\ &\Longrightarrow \exists d_1,\ldots,d_n \bigl( f(a)=d_1\lessdot\cdots\lessdot d_n=b \bigr)\\ &\Longrightarrow a=c_1\lessdot\cdots\lessdot c_n=b, \end{align} where $c_i=f^{-1}(d_i)$. As $a\lessdot b$, it follows that $n=2$ and $b=c_2$, and therefore, $f(b)=d_2$, whence $f(a)\lessdot f(b)$.
For the converse implication use $f^{-1}$ (which satisfies $c\leq d$ iff $f^{-1}(c)\leq f^{-1}(d)$).
Suppose now that $f:P\to Q$ is a bijection satisfying $a\lessdot b$ iff $f(a)\lessdot f(b)$.
If $a\leq b$, then there exist $c_1,\ldots,c_n$ such that $$a=c_1\lessdot\cdots\lessdot c_n=b,$$ and so $$f(a)=f(c_1)\lessdot\cdots\lessdot f(c_n)=f(b),$$ yielding $f(a)\leq f(b)$. Again, for the converse, use $f^{-1}$.