Short answer
What you say is true for steady conditions, in conductors with uniform resistivity and section, for the average of the electric field on the sections of the conductors.
Details
The answer, and the answers given by other members, should be clear if we consider the constitutive equation linking the electric field $\mathbf{e}$ and the current density $\mathbf{j}$ through the resistivity $r$ (or its inverse, the conductivity $\sigma = \frac{1}{r})$,
$\mathbf{e} = r \mathbf{j} \qquad , \qquad \mathbf{j} = \sigma \mathbf{e}$.
- perfect conductors have zero resistivity, $r = 0$, and thus $\mathbf{e} = \mathbf{0}$ inside the conductor for every value of the current;
- in non-perfect conductors wit non-zero electric current density $\mathbf{j}$, the electric field is not equal to zero, but it's equal to $\mathbf{e} = r \mathbf{j}$.
Steady conditions for conductors with constant section.
Now, let's consider here the steady conditions, and take a volume of the conductor. Electric charges can leave the volume through its lateral surface, otherwise electric charge would accumulate on the surface of the conductor, or leave the conductor; the integral balance equation of electric charge reads,
$0 =
\displaystyle \int_{S_1} \mathbf{j} \cdot \mathbf{\hat{n}}_1 + \int_{S_2} \mathbf{j} \cdot \mathbf{\hat{n}}_2 =
- \int_{S_1} \mathbf{j} \cdot \mathbf{\hat{t}}_1 + \int_{S_2} \mathbf{j} \cdot \mathbf{\hat{t}}_2 \qquad\rightarrow \qquad \int_{S_1} \mathbf{j} \cdot \mathbf{\hat{t}}_1 = \int_{S_2} \mathbf{j} \cdot \mathbf{\hat{t}}_2$,
being $S_1$, $S_2$ two sections of the wire, with the respective normal unit-vector pointing outwards the volume, and $\mathbf{\hat{t}}$ is the unit vector "pointing always in the same direction when you move along the conductor", $\mathbf{\hat{n}}_1 = -\mathbf{\hat{t}}_1 $, $\mathbf{\hat{n}}_2 = \mathbf{\hat{t}}_2 $.
Using the constitutive law $\mathbf{j} = \sigma \mathbf{e}$, if the conductivity is uniform and equal on the two sections $\sigma|_{S_1} = \sigma|_{S_2} = \sigma$
$ \displaystyle \sigma \int_{S_1} \mathbf{e} \cdot \mathbf{\hat{t}}_1 = \sigma \int_{S_2} \mathbf{e} \cdot \mathbf{\hat{t}}_2$
and thus the average flux of the electric field across each section of the conductor with the same resistivity has the same value.
Far from any bend in the circuit, for symmetry considerations, the current density, and thus the electric field has the same direction as the axis of the conductor and thus,
$e_1 A_1 = e_2 A_2 \quad \rightarrow (A_1 = A_2) \rightarrow \qquad e_1 = e_2$.
Close to circuit bends, the electric field could be non uniform in space, and the relations hold only with the average quantities.
Circuit approximation. With the circuit approximation, lumping the conductor in a line with section as a property, we take the average value of a physical quantity on a section as the uniform value on that section, and both the current density and the electric field aligned with the unit vector tangent to the axis of the conductor,
$\displaystyle \mathbf{e} = e \, \mathbf{\hat{t}} = \dfrac{1}{A} \int_A \mathbf{e} \cdot \mathbf{\hat{t}} dA \, \mathbf{\hat{t}}$
$\displaystyle \mathbf{j} = j \, \mathbf{\hat{t}} = \dfrac{1}{A} \int_A \mathbf{j} \cdot \mathbf{\hat{t}} dA \, \mathbf{\hat{t}}$,
so that the relations obtained before holds $e_1 = e_2$.