Point (1) is correctly addressed in the comment by Hoot. As for point (2), your intuition is on the right track. On the other hand, you should keep track of the multiplicities of the loci involved.
As a matter of example, let $X$ be the projective plane, $Y$ the cuspidal rational curve, and $Z$ the singular point of $Y$. $Z$ is a regular subvariety of $X$, so the exceptional divisor is just a copy of $\mathbb{P}^1$ (in general, if you blow up something singular, the exceptional locus might be pretty ugly though). The strict transform of $Y$ (i.e. $Y'$ in your notation) is going to be a smooth rational curve tangent to $E$. This reflects that $Y$ has multiplicity 2 along $Z$. This gives you that $\pi^*Y= Y'+2E$. As you see, the ingredients are exactly the ones you expected, but, in this case, they are weighted with coefficients depending on the singularities of $Y$ along $Z$.
Edit I am reading your answer more carefully now. If both $Y$ and $Z$ are smooth, then claim (2) is fine as well.
Addendum You comment is right. The blow up is an isomorphism over $X \setminus Z$. In particular, if $\widehat{Y}$ is disjoint from $Z$, its strict transform $\widehat{Y}'$ coincides with the pullback $\pi^*(\widehat{Y})$ , and it is isomorphic to $\widehat{Y}$. Now, if $Y$ and $\widehat{Y}$ are linearly equivalent, so are their pullbacks (just because the isomorphism between $\mathcal{O}_X(Y)$ and $\mathcal{O}_X(\widehat{Y})$ induces an isomorphism between their pullbacks). On the other hand, this is telling you that the strict transforms of linear equivalent divisors are not linearly equivalent if just one of the two goes through $Z$.
Let me be more explicit. Blow up a point $P$ in $\mathbb{P}^2$. Let $L_1$ be a line through $P$, and $L_2$ a line not containing $P$. Denote by $M_1$ and $M_2$ the respective strict transforms. Then, by what above said, we have $\pi^*L_1=M_1+E$, and $\pi^*L_2=M_2$. By Bezout theorem we know that the intersection products $L_1 \cdot L_2=(L_1)^2=(L_2)^2=1$. In particular $L_1$ and $L_2$ meet properly at one point, say $Q$. Now, since $L_2$ does not go through $P$, we have that the pullbacks $M_1+E$ and $M_2$ meet properly at one point (the only premiage of $Q$). Given that these divisors are also linearly equivalent to each other, we get $1=(M_2)^2=M_2 \cdot (M_1+E)=(M_1+E)^2$. In particular, we get $1=(M_1+E)^2=M_1^2+2M_1\cdot E+ E^2$. Since $M_1$ and $E$ meet properly at one point, we know $M_1 \cdot E=1$. Then, we know that $E^2=\mathrm{deg}\mathcal{O}_{X'}(E)_{|E}$. By the description in sections 7 and 8 in chapter 2 of Hartshorne, we know that this is the relative $\mathcal{O}(1)$ bundle, that $E=\mathbb{P}^1$; these things together tell us that $\mathcal{O}_{X'}(E)_{|E}\cong \mathcal{O}_{\mathbb{P}^1}(-1)$. Thus, that degree is $-1$, so $E^2=-1$. This is negative self intersection is phrased as $E$ does not deform: there is no other effective divisor equivalent to $E$. Now, putting this in our previous equation, we get that $(M_1)^2=0$. As you see, $(M_1)^2 \neq (M_2)^2$; in particular, they can not be linearly equivalent.
This MO post is a great reference if you're ever trying to figure out or remember when for a morphism $f:X\to Y$ we will have $f_*\mathcal{O}_X=\mathcal{O}_Y$.
Here's the relevant portion of that answer for this post, in order to make this answer self-contained:
The case of an arbitrary projective morphism.
Now when $f:X\to Y$ is any projective morphism, then $f_*\mathscr O_X$ is a coherent
$\mathscr O_Y$-module, hence we get a factorization of $f$ as $h\circ g:X\to Z\to Y$,
where $h:Z\to Y$ is affine, and where also $h_*(\mathscr O_Z) = f_*\mathscr O_X$.
Then $h$ is not only an affine map, but since $h_*(\mathscr O_Z)$ is a coherent $\mathscr
O_Y$-module, $h$ is also a finite map. Moreover $g:X\to Z$ is also projective and since
$g_*(\mathscr O_X) = \mathscr O_Z$, it can be shown that the fibers of $g$ are connected.
Hence an arbitrary projective map $f$ factors through a projective map $g$ with connected
fibers, followed by a finite map $h$. Thus in this case, the algebra $f_*\mathscr O_X$
determines exactly the finite part $h:Z\to Y$ of $f$, whose points over $y$ are precisely
the connected components of the fiber $f^{-1}(y)$.
One corollary of this is "Zariski's connectedness theorem". If $f:X\to Y$ is projective
and birational, and $Y$ is normal then $f_*\mathscr O_X= \mathscr O_Y$, and all fibers
of $f$ are connected, since in this case $Z = Y$ in the Stein factorization described
above. If we assume in addition that $f$ is quasi finite, i.e. has finite fibers, then
$f$ is an isomorphism. More generally, if $Y$ is normal and $f:X\to Y$ is any birational,
quasi - finite, morphism, then $f$ is an embedding onto an open subset of $Y$ ("Zariski's
'main theorem' "). More generally still, any quasi finite morphism factors through
an open embedding and a finite morphism.
This applies to your situation as follows: the blowup map $\pi:\widetilde{X}\to X$ is a projective birational map with connected fibers. Since projective and connected fibers are preserved under base change, we see that the base change of this map $E\to Y$ is again projective with connected fibers, so we may also apply the result there, via the Stein factorization described in the first paragraph (even though this last morphism is not birational - $\dim E=\dim X-1\neq \dim Y$).
Best Answer
By Grothendieck duality $$ Rb_*\mathcal{O}_Y(K_Y + E) \cong Rb_*R\mathcal{H}\mathit{om}(\mathcal{O}_Y(-E),\omega_Y) \cong Rb_*R\mathcal{H}\mathit{om}(\mathcal{O}_Y(-E),b^!\omega_X) \cong R\mathcal{H}\mathit{om}(Rb_*(\mathcal{O}_Y(-E)),\omega_X) \cong I_Z^\vee \otimes \omega_X, $$ where $(-)^\vee$ stands for the derived dual.