Denote with
$$
M(r, f) = \max\{ |f(z)| : |z| \le r \}
$$
the maximum of $|f|$ on disks of radius $r$. It follows from Cauchy's integral formula for the derivative that
$$
M(r, f') \le M(r+1, f) \, ,
$$
which implies $\lambda(f') \le \lambda(f)$. (For the details, see for example $f$ is entire, show that the order of $f'$ is less than or equal to that of $f$ .)
For the other direction note that
$$
f(r e^{i \phi}) = f(0) + \int_0^r f'(t e^{i \phi})e^{i \phi} \, dt
$$
implies
$$
M(r, f) \le |f(0)| + r M(r, f') \, .
$$
from which $\lambda(f) \le \lambda(f')$ follows in a similar way.
The answer is $1$ as follows; first it is enough to find $g$ (entire non zero, order $1$) nonnegative and integrable on the real line, so $M=\int_{-\infty}^{\infty}g(x)dx < \infty, g(x)\ge 0, x \in \mathbb R$ as then $f(z)=\int_{[0,z]}g(t)dt$ works, since for $x<y, f(y)-f(x)=\int_x^yg(t)dt$ and $g(t)>0$ except possible at countably many discrete zeroes so in particular on some small interval in $(x,y)$ while clearly $-M < f(x) <M$, $f$ is entire and of order $1$ by general theory
But now pick any nonzero $h \in L^2(-a,a)$ with some $a>0$ and by Paley Wiener (very easy part - converse is the difficult part) $G(z)=\int_{-a}^ah(t)e^{izt}dt$ is an entire (nonzero) exponential function (order $1$ functions are also called exponential for obvious reasons) of type $a$ which is square integrable on any horizontal line, $\int_{-\infty}^{\infty}|G(x+iy)|^2dx < \infty$ for any fixed $y \in \mathbb R$.
If we take $h$ odd, then $G$ is conjugate invariant since $$G(x) =\int_{-a}^ah(t)(\cos xt+i\sin xt)=\int_{-a}^ah(t)\cos xtdt, x \in \mathbb R$$
Then if we take $g(z)=G(z)\overline {G(\bar z)}$ (or $g(z)=G^2(z)$ if $G$ is conjugate invariant), one clearly has $g$ entire non zero, $g(x) \ge 0$ for real $x$ and $\int_{-\infty}^{\infty}g(x)dx=\int_{-\infty}^{\infty}|G(x)|^2dx< \infty$, while $g$ is of order $1$ and type $2a$ by general theory
(note that if one wants, one can obtain a strictly positive $g$ by an elaboration of the above, picking a $y$ where $G$ has no zeroes on the horizontal line $x+iy$ and defining $G_1(z)=G(z+iy)$ and using the square integrability of $G$ on the horizontal line passing through $iy$) etc)
Edit later; choose $a=1, h(t)=t/|t|$ the signum of $t \ne 0$ so $G(z)=2\int_0^1\cos ztdt=\frac{2}{z}\sin z$ so $g(z)=4\frac{\sin^2(z)}{z^2}$ is the example of @reuns in the comment above
Best Answer
Without loss of generality, $\lambda_1 < \lambda_2$. You already showed that the order $\lambda $ of $f$ is at most $ \max (\lambda_1,\lambda_2) = \lambda_2$.
By the same argument, applied to $f_2 = f + (-f_1)$, is $$ \lambda_2 \le \max(\lambda, \lambda_1) = \lambda $$ since the maximum can not be $\lambda_1$.
Therefore $\lambda = \lambda_2 = \max (\lambda_1,\lambda_2)$.