Is there a simple and explicit continuous function $f\colon[0,\infty)^2\to\mathbb C$ such that $f$ is analytic on $(0,\infty)^2$ and $|f(x+iy)|/(x+y)\to1$ as $x+y\to\infty$, where $(x,y)\in[0,\infty)^2$?
It seems not too hard to construct such a function $f$ as the sum of a series of polynomials, as is done in the proof of the Carleman approximation theorem; cf. [1], [2]. However, that construction is far from explicit.
Such function does not exist, "explicit" or not. Consider $u(z)=\log|f(z)/z|$. Your condition implies that $u$ is harmonic and bounded when $|z|>r$ for some $r>0$ and $z$ is in the first quadrant. Moreover it tends to $0$ on positive real and imaginary axes. But on the line $x=y$ it tends to $\log\sqrt{2}>0$. This contradicts the Phragmen--Lindelof Principle.
EDIT. If $x=y$ then $$u(x+ix)=\log|f(x+ix)/(x+ix)|=\log( 2x/(x\sqrt{2}))+o(1)\to\log\sqrt{2}.$$ The form of Phragmen-Lindelof that is needed is the following: Let $u$ be harmonic in a region $D$, $u$ is bounded from above and $a\in\partial D$. If $$\limsup_{z\to\zeta}u(z)\leq C,$$ for all $\zeta\in\partial D\cap V$, where $V$ is a punctured neighborhood of $a$, then $$\limsup_{z\to a}u(z)\leq C.$$ In our case $a=\infty$.
Because Alexandre's solution uses a version of the Phragmén–-Lindelöf principle (PLP) that requires some extra effort to find or prove, I would like to present the following modification of his answer, which uses only a well-known version of the PLP and proceeds without taking the logarithm:
Let $g(z):=f(z)/(z+b(1+i))$, where $b>0$ is large enough to ensure that $|g|\le1.1$ on the entire positive real and imaginary semi-axes. Then $g$ is continuous and bounded on $[0,\infty)^2$ and analytic on $(0,\infty)^2$. So, by the PLP found in Section 5.6 of [1], $|g|\le1.1$ on $[0,\infty)^2$, which contradicts the fact that $|g(x+ix)|\to\sqrt2$ as $x\to\infty$.
As for the version of the PLP stated and used by Alexandre, it seems interesting and apparently not hard to prove indeed. Indeed, by the Riemann mapping theorem, without loss of generality the region $D$ is the open unit disk. One can now use a tilting idea similar to using $b$ in the above solution or to the tilting used in proofs of the PLP itself. Namely, let $u_n(z):=u(z)+n\ln|z|$ for $n>0$. Then $u_n$ is harmonic and $u_n=u$ on the boundary $\partial D$. Moreover, for any $\epsilon>0$ and an appropriate small enough neighborhood $U$ of $a$ such that $U\subset V$, one can choose $n>0$ large enough so that $\limsup_{z\to w}u_n(z)\le C+\epsilon$ for all $w\in\partial(U\cap D)\setminus\{a\}$; the progress made here is that now $w$ does not have to lie on $\partial D$. Then, by an "appropriate" version of the PLP (if such exists and which I have not been able to find either), $u_n\le C+\epsilon$ on $U\cap D$, whence $\limsup_{z\to a}u(z)=\limsup_{z\to a}u_n(z)\le C+\epsilon$, for any $\epsilon>0$.
