Given a Hilbert space $H$, let us build another Hilbert space $G$ by taking each vector $\xi $ in $H$ and "painting it
green". By this I mean that we are disguising $\xi $ in any way we want, changhing its nature, but keeping track of
which vector $\xi $ we started with. If you cannot imagine a green vector, you could alternatively think of the ordered
pair $(\xi ,*)$, where $*$ is anything you like.
The set $G$ of all green vectors is then given the structure of a Hilbert space in the most natural way possible. For
instance if $x$ and $y$ are a green vectors, we define $x+y$ as follows:
scratch the paint of $x$ and $y$, resulting in vectors $\xi $ and $\eta $ in $H$, respectively,
add $\xi $ and $\eta $, as vectors in $H$, and call the sum $\zeta $,
paint $\zeta $ green and call the resulting vector $z$,
set $x+y:= z$.
We may then define a unitary operator $U:G\to H$ by seting
$$
U(\xi ) = \xi \text{ painted green}, \quad \forall \xi \in H.
$$
If $T$ is any bounded operator on $H$, one can likewise turn it into an operator $S$ acting on $G$ via:
given $x$ in $G$,
scratch the paint of $x$, resulting in the vector $\xi $ in $H$,
apply $T$ to $\xi $, resulting in the vector $\eta $ in $H$,
paint $\eta $ green, resulting in the vector $y\in G$,
set $S(x)=y$.
Notice that the whole algorithm defining $S$ above may be summarized by
$$
S = UTU^{-1}.
$$
The moral of the story is that when two operators are conjugate, they should be each viewed as a disguise of the other,
and hence they have exactly the same properties, as long as these properties are defined purely in the language of
Functional Analysis.
Best Answer
Yes, it can happen. The canonical example is the unilateral shift. For instance take $H=\ell^2(\mathbb N)$, and put $$ Ax=(0,x_1,x_2,x_3,\ldots). $$ It is straightforward to see that $A$ is an isometry, which immediately implies that $A^\dagger A=1$. But $A$ is not surjective. With just a tiny more work, one can see that $$ AA^\dagger x=(0,x_2,x_3,\ldots). $$ Such an operator can be defined in any separable infinite-dimensional Hilbert space, by fixing an orthonormal basis $\{e_n\}$ and defining $A$ as the linear operator that satisfies $Ae_n=e_{n+1}$ for all $n$.
As mentioned by David, the Wold Decomposition guarantees that the only counterexamples are versions of this one.